Mastery Guide: Tangent to a Circle, Touching Circles, Inscribed Angles & Cyclic Quadrilaterals
📖 1. Unit Overview & Target Learning Outcomes
Unit 10 establishes the foundational theorems of differential and Euclidean geometry regarding tangents, circles touching internally/externally, and cyclic quadrilaterals. By mastering this unit, students will be able to:
- Define & Identify Tangents: Distinguish rigorously between a secant line (2 intersection points) and a tangent line (exactly 1 point of contact).
- Prove Radial Orthogonality (Theorems 10.1 & 10.2): Prove that the tangent line is perpendicular to the radial segment at the point of contact ($m\angle = 90^\circ$).
- Compute Tangents from External Points (Theorem 10.3): Prove that two tangents drawn from an external point to a circle are congruent ($AB = AC$).
- Analyze Touching Circles (Theorems 10.4 & 10.5): Determine the distance between centres of circles touching externally ($d = r_1 + r_2$) and internally ($d = r_1 - r_2$).
- Apply the Alternate Segment Theorem (Theorem 10.6): Prove and calculate angles between tangents and chords intersecting the alternate arc.
- Relate Central & Inscribed Angles (Theorem 10.7 & 10.8): Establish that central angles are double inscribed angles, and angles in the same segment are equal.
- Master Semicircle & Segment Angle Bounds (Theorems 10.9, 10.10, 10.11): Deduce Thales's Theorem ($90^\circ$ in a semicircle), acute angles in major segments, and obtuse angles in minor segments.
- Solve Cyclic Quadrilateral Properties (Theorem 10.12 & Ptolemy's Theorem): Prove opposite angles are supplementary ($180^\circ$) and apply Ptolemy's diagonal-product theorem ($p \cdot q = ac + bd$).
💡 2. Kid-Friendly Tips & Memory Hooks
🎯 The "T-Square" Tangent Rule
Whenever a radius meets a tangent at the point of contact, it forms an infallible $90^\circ$ right angle. Always draw the little square box and immediately unlock Pythagoras' Theorem ($a^2 + b^2 = c^2$)!
🍦 The Ice Cream Cone Hook
Two tangents from an external point look just like an ice cream cone! The two waffle edges (tangents $AB$ and $AC$) are always exactly equal in length ($AB = AC$).
🏹 The Bow & Arrow Rule
The central angle at the center of the circle is always twice as big as the angle pulled back at the boundary circumference ($\text{Center} = 2 \times \text{Circumference}$).
🌍 3. Real-World Connections & Engineering Applications
- Bicycle Chains & Belt Pulleys: The drive belt or chain leaves the gears along common tangent lines. The distance between the gear centres determines required belt lengths using $d = r_1 + r_2$ or trigonometric tangent projections.
- Orbital Trajectories & Spacecraft Re-entry: When a rocket escapes planetary orbit, its escape path is along the instantaneous tangent vector perpendicular to the gravitational radial vector.
- Architecture & Circular Arches: Gothic and Roman arched windows utilize intersecting circular segments where stone load distributions rely on cyclic quadrilateral equilibrium ($180^\circ$ supplementary load angles).
- Wheel Rolling Dynamics: A circular wheel on flat road touches at a single instantaneous point of contact, ensuring pure rolling friction without slip.
🔑 4. Study Cues & Essential Conceptual Inquiries
- Why can a circle have only one tangent at a specific boundary point, but two tangents from an external point? (At the boundary point, the normal radial vector is unique, allowing only one perpendicular line. Outside, two lines of sight touch the circle.)
- Why is the angle in a semicircle always $90^\circ$ regardless of where the point is on the arc? (Because the central angle is a straight line $180^\circ$, and by Theorem 10.7, half of $180^\circ$ is identically $90^\circ$.)
- How does the Alternate Segment Theorem simplify complex chord-tangent angle calculations in board exams? (It replaces complex trigonometry with direct angle equivalence: $\angle \text{Tangent-Chord} = \angle \text{Alternate Inscribed Angle}$.)
🌟 5. Section-by-Section Theoretical Foundations
Key Geometric Definitions & Properties:
| Geometric Concept | Definition & Core Property | Mathematical Formula |
|---|---|---|
| Tangent Line | A line that touches a circle at exactly one point (Point of Contact). | $AB \perp OB$ at point of contact $B$. |
| Secant Line | A line that cuts through a circle at two distinct boundary points. | Intersects circle at points $S$ and $T$. |
| External Tangents | Two equal tangent segments drawn from an external point $A$. | $AB = AC$ (Theorem 10.3). |
| Externally Touching Circles | Two circles touching at one exterior point; distance between centres is $r_1 + r_2$. | $d = r_1 + r_2$ (Theorem 10.4). |
| Internally Touching Circles | One circle inside another touching at one point; distance between centres is $r_1 - r_2$. | $d = r_1 - r_2$ (Theorem 10.5). |
| Cyclic Quadrilateral | A 4-sided polygon whose 4 vertices all lie on a single circle. | Opposite angles are supplementary: $\angle A + \angle C = 180^\circ$. |
📐 6. Comprehensive Theorem Repository (Theorems 10.1 to 10.12)
Below are the 12 complete, formal FBISE board examination proofs with official book figures, SVG vector diagrams, Statements, and Reasons:
Theorem 10.1 • Perpendicular to Radial Segment is a Tangent
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| In right-angled triangle $\triangle ABC$, $m\angle ABC = 90^\circ$. | Given ($PQ \perp AB$). |
| The line segment $AC$ is the hypotenuse of $\triangle ABC$. | Side opposite to right angle is hypotenuse. |
| Therefore, $AC > AB$. | In a right triangle, hypotenuse is longer than any leg. |
| Since $AB$ is the radius of the circle, point $C$ lies outside the circle. | Distance $AC > \text{radius } AB$. |
| Similarly, every point on line $PQ$ except $B$ lies strictly outside the circle. | Perpendicular distance is the shortest distance. |
| Hence, line $PQ$ intersects the circle at only one point $B$. | Only point $B$ is on the circumference. |
| Therefore, line $PQ$ is tangent to the circle at point $B$. | Definition of a tangent line. |
Theorem 10.2 • Tangent is Perpendicular to Radial Segment
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Line $PQ$ is tangent to the circle at point $B$. | Given. |
| Therefore, point $B$ lies on the circle and every other point $C$ on $PQ$ lies outside the circle. | Definition of tangent line. |
| Thus, distance $AC > AB$ for every point $C \neq B$ on $PQ$. | $AB = \text{radius}$, $C$ is external. |
| The segment $AB$ is the shortest distance from centre $A$ to line $PQ$. | $AB < AC$ for all points $C$. |
| The shortest segment from a point to a line is the perpendicular segment. | Geometric axiom of perpendicular distance. |
| Hence, $AB \perp PQ$ ($m\angle ABQ = 90^\circ$). | Shortest line is perpendicular. |
Theorem 10.3 • Tangents from an External Point are Equal
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| In $\triangle OBA$ and $\triangle OCA$: | Comparing the two right triangles. |
| $m\angle OBA = m\angle OCA = 90^\circ$ | Radius $\perp$ Tangent at contact points $B$ and $C$ (Theorem 10.2). |
| $\text{Hypotenuse } OA = \text{Hypotenuse } OA$ | Common side. |
| $OB = OC$ | Radii of the same circle ($r$). |
| $\triangle OBA \cong \triangle OCA$ | R.H.S. Congruence Postulate. |
| Hence, $AB = AC$. | Corresponding sides of congruent triangles. |
Theorem 10.4 • Centres of Externally Touching Circles
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Let the common tangent to both circles at contact point $C$ be line $PQ$. | Construction. |
| $AC \perp PQ$ | Radial segment $AC \perp$ tangent $PQ$ at contact point $C$ (Theorem 10.2). |
| $BC \perp PQ$ | Radial segment $BC \perp$ tangent $PQ$ at contact point $C$ (Theorem 10.2). |
| Since $AC$ and $BC$ are both perpendicular to line $PQ$ at the same point $C$, the points $A, C, B$ are collinear. | Only one perpendicular can be drawn to a line at a given point. |
| Therefore, distance $AB = AC + BC$. | Segment addition postulate for collinear points. |
| Hence, $AB = r_1 + r_2$. | Since $AC = r_1$ and $BC = r_2$. |
Theorem 10.5 • Centres of Internally Touching Circles
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Let line $PQ$ be the common tangent at point $C$. | Construction. |
| $AC \perp PQ$ and $BC \perp PQ$ | Radius $\perp$ Tangent at point of contact (Theorem 10.2). |
| Therefore, points $A, B$, and $C$ lie on the same straight line. | Perpendiculars at same point $C$ are collinear. |
| For internally touching circles, centre $B$ lies on segment $AC$. | $r_1 = AC > r_2 = BC$. |
| Distance $AB = AC - BC$. | Segment subtraction postulate. |
| Hence, $AB = r_1 - r_2$. | Since $AC = r_1$ and $BC = r_2$. |
Theorem 10.6 • Alternate Segment Theorem
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| In right triangle $\triangle TDB$: | Diameter $TD$ subtends $90^\circ$ at circle. |
| $m\angle TBD = 90^\circ$ | Angle in a semicircle is a right angle (Theorem 10.9). |
| $m\angle BTD + m\angle TDB = 90^\circ$ ...(i) | Acute angles of a right-angled triangle sum to $90^\circ$. |
| Also, $TD \perp XY \implies m\angle XTD = 90^\circ$. | Radius/Diameter $\perp$ Tangent at contact point $T$. |
| $m\angle XTP + m\angle PTD = 90^\circ$ ...(ii) | Complementary angles. |
| From (i) and (ii): $m\angle XTP = m\angle TDB$. | Both are equal to $90^\circ - m\angle BTD$. |
| But $m\angle TDB = m\angle PBT$. | Angles in the same segment subtended by chord $PT$ (Theorem 10.8). |
| Hence, $m\angle XTP = m\angle PBT$. | Substitution of equal angle measures. |
Theorem 10.7 • Central Angle is Double the Inscribed Angle
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| In $\triangle OCA$, $OA = OC$ (radii) $\implies m\angle 1 = m\angle 3$. | Base angles of isosceles triangle are equal. |
| Exterior angle $m\angle AOC_1 = m\angle 1 + m\angle 3 = 2m\angle 1$. ...(i) | Exterior angle of a triangle equals sum of opposite interior angles. |
| In $\triangle OCB$, $OB = OC$ (radii) $\implies m\angle 2 = m\angle 4$. | Base angles of isosceles triangle are equal. |
| Exterior angle $m\angle BOC_1 = m\angle 2 + m\angle 4 = 2m\angle 2$. ...(ii) | Exterior angle property. |
| Adding (i) and (ii): $m\angle AOC_1 + m\angle BOC_1 = 2(m\angle 1 + m\angle 2)$. | Addition of equations. |
| Hence, $m\angle AOB = 2m\angle ACB$. | Since $\angle AOB = \angle AOC_1 + \angle BOC_1$ and $\angle ACB = \angle 1 + \angle 2$. |
Theorem 10.8 • Angles in the Same Segment are Equal
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Minor arc $AB$ subtends central angle $\angle AOB$. | Arc subtends angle at centre $O$. |
| $m\angle AOB = 2m\angle ACB$ ...(i) | Central angle is double inscribed angle $\angle ACB$ (Theorem 10.7). |
| $m\angle AOB = 2m\angle ADB$ ...(ii) | Central angle is double inscribed angle $\angle ADB$ (Theorem 10.7). |
| From (i) and (ii): $2m\angle ACB = 2m\angle ADB$. | Both equal central angle $m\angle AOB$. |
| Dividing by 2: $m\angle ACB = m\angle ADB$. | Angles in the same segment are equal. |
Theorem 10.9 • Angle in a Semicircle is a Right Angle (Thales's Theorem)
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Diameter $AB$ is a straight line segment passing through centre $O$. | Definition of diameter. |
| Therefore, the central angle $m\angle AOB = 180^\circ$. | Angle of a straight line. |
| By Theorem 10.7, $m\angle AOB = 2m\angle APB$. | Central angle is double inscribed angle. |
| $180^\circ = 2m\angle APB$. | Substitution. |
| Hence, $m\angle APB = \frac{180^\circ}{2} = 90^\circ$. | Right angle. |
Theorem 10.10 • Angle Inscribed in a Major Segment is Acute
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Since chord $AB$ subtends a minor arc at the center, central angle $m\angle AOB < 180^\circ$. | Minor arc spans strictly less than a straight semicircle ($180^\circ$). |
| By Theorem 10.7, $m\angle AOB = 2m\angle APB$. | Central angle is double the inscribed angle on major arc. |
| $2m\angle APB < 180^\circ$. | Substitution of $m\angle AOB < 180^\circ$. |
| Dividing both sides by 2: $m\angle APB < \frac{180^\circ}{2} = 90^\circ$. | Division property of inequalities. |
| Hence, the angle in a major segment is an acute angle ($< 90^\circ$). | An angle strictly between $0^\circ$ and $90^\circ$ is acute. |
Theorem 10.11 • Angle Inscribed in a Minor Segment is Obtuse
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| The major arc corresponding to minor chord $AB$ subtends reflex central angle $\text{Reflex } m\angle AOB > 180^\circ$. | A major arc central rotation strictly exceeds a straight angle ($180^\circ$). |
| By Theorem 10.7, $\text{Reflex } m\angle AOB = 2m\angle AQB$. | Central angle is double the inscribed angle in the minor segment. |
| $2m\angle AQB > 180^\circ$. | Substitution of $\text{Reflex } m\angle AOB > 180^\circ$. |
| Dividing both sides by 2: $m\angle AQB > \frac{180^\circ}{2} = 90^\circ$. | Division property of inequalities. |
| Since $90^\circ < m\angle AQB < 180^\circ$, $\angle AQB$ is an obtuse angle. | Definition of an obtuse angle. |
Theorem 10.12 • Opposite Angles of a Cyclic Quadrilateral are Supplementary
Official FBISE ProofFormal 2-Column Proof (Statements & Reasons):
| Statements | Reasons |
|---|---|
| Arc $BCD$ subtends central angle $\angle BOD$ and inscribed angle $\angle BAD$. | Arc subtends angles at centre and circumference. |
| Central angle $m\angle BOD = 2m\angle BAD$ ...(i) | Central angle is double inscribed angle (Theorem 10.7). |
| Arc $BAD$ subtends reflex central angle $\text{Reflex } \angle BOD$ and inscribed angle $\angle BCD$. | Opposite major arc subtends reflex central angle. |
| Reflex central angle $m(\text{Reflex } \angle BOD) = 2m\angle BCD$ ...(ii) | Theorem 10.7 applied to reflex arc. |
| Adding (i) and (ii): $m\angle BOD + m(\text{Reflex } \angle BOD) = 2(m\angle BAD + m\angle BCD)$. | Addition of equations. |
| Since $m\angle BOD + m(\text{Reflex } \angle BOD) = 360^\circ$ (complete circular angle): | Total angular rotation around centre is $360^\circ$. |
| $360^\circ = 2(m\angle A + m\angle C) \implies m\angle A + m\angle C = 180^\circ$. | Dividing by 2. |
| Similarly, $m\angle B + m\angle D = 180^\circ$. | Sum of all four angles in quadrilateral is $360^\circ$. |
🎯 7. Unit Synthesis Summary
Chapter 10 provides a complete geometric synthesis connecting line tangency, circular touch conditions, and angular theorems:
- Orthogonality: Tangents are everywhere perpendicular to radii at contact points, establishing right-angled relations for Euclidean calculation.
- Congruence: Pairs of tangents from an external point are congruent ($AB = AC$), creating symmetric isosceles triangles and right-angled pairs.
- Collinearity: Touching circles align their centres and common contact point along a single collinear axis, giving $d = r_1 \pm r_2$.
- Angular Invariance: Central angles strictly double inscribed angles, angles in the same segment remain identical, and cyclic quadrilaterals preserve exact $180^\circ$ opposite angular balance.
📝 Part 2: Solved Textbook Exercises (Step-by-Step Manual)
Exercise 10.1 • Tangents, Radial Perpendicularity & Right Triangles
Exhaustive solutions for all 15 questions & subparts (Pages 220–222)
Exercise 10.1 - Q1(1(i)) Tangents & Radial Orthogonality: In the right-angled triangle formed by a tangent and radial segment, the hypotenuse is $13\text{ cm}$ and one leg is $12\text{ cm}$. Find the unknown radius $x$.
Step 1: Apply the Radius-Tangent Perpendicularity Theorem (Theorem 10.2):
The radial segment joining the center to the point of contact is perpendicular to the tangent line. Thus, the triangle is a right-angled triangle with hypotenuse $c = 13\text{ cm}$ and leg $b = 12\text{ cm}$.
Step 2: Apply the Pythagorean Theorem:
$$x^2 + 12^2 = 13^2$$
$$x^2 + 144 = 169 \implies x^2 = 169 - 144 = 25$$
$$x = \sqrt{25} = 5\text{ cm}$$
Final Answer:
$$\mathbf{x = 5\text{ cm}}$$
Exercise 10.1 - Q1(1(ii)) Tangents & Radial Orthogonality: A tangent of length $2.1\text{ cm}$ is drawn to a circle of radius $2\text{ cm}$. Find the distance $d$ from the center of the circle to the external point.
Step 1: Set up the right triangle relations:
Radius $r = 2\text{ cm}$, Tangent length $t = 2.1\text{ cm}$. The distance from the center to the external point is the hypotenuse $d$.
Step 2: Calculate hypotenuse using Pythagoras Theorem:
$$d = \sqrt{r^2 + t^2} = \sqrt{2^2 + (2.1)^2} = \sqrt{4 + 4.41} = \sqrt{8.41} = 2.9\text{ cm}$$
Final Answer:
$$\mathbf{d = 2.9\text{ cm}}$$
Exercise 10.1 - Q1(1(iii)) Tangents & Radial Orthogonality: A tangent is drawn from an external point $P$ at a distance of $17\text{ cm}$ from the center of a circle of radius $8\text{ cm}$. Find the length of the tangent $x$.
Step 1: Set up the Pythagorean equation:
Hypotenuse $OP = 17\text{ cm}$, Radius $r = 8\text{ cm}$, Tangent $x$.
$$x = \sqrt{OP^2 - r^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\text{ cm}$$
Final Answer:
$$\mathbf{x = 15\text{ cm}}$$
Exercise 10.1 - Q1(1(iv)) Tangents & Radial Orthogonality: A tangent of length $6\sqrt{2}\text{ cm}$ is drawn to a circle of radius $6\text{ cm}$. Find the distance from the center of the circle to the external point.
Step 1: Apply the Pythagorean Theorem:
$$d = \sqrt{r^2 + t^2} = \sqrt{6^2 + (6\sqrt{2})^2} = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}$$
Final Answer:
$$\mathbf{d = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}}$$
Exercise 10.1 - Q1(1(v)) Tangents & Radial Orthogonality: Two tangents drawn from an external point to a circle contain an angle of $50^\circ$. Find the central angle subtended by the line segments joining the points of contact to the center.
Step 1: Apply the Quadrilateral Angle Sum Property:
In quadrilateral $OBAC$ formed by center $O$, external point $A$, and contact points $B, C$:
$\angle B = 90^\circ$ and $\angle C = 90^\circ$ (Theorem 10.2).
$$\angle BOC + \angle BAC + 90^\circ + 90^\circ = 360^\circ$$
$$\angle BOC + 50^\circ + 180^\circ = 360^\circ \implies \angle BOC = 360^\circ - 230^\circ = 130^\circ$$
Final Answer:
$$\mathbf{\angle BOC = 130^\circ}$$
Exercise 10.1 - Q1(1(vi)) Tangents & Radial Orthogonality: The central angle subtended by two radii to tangent contact points is $110^\circ$. Find the angle between the two tangents drawn from the external point.
Step 1: Use the supplementary property of tangent-radius angles:
The angle between the tangents $\theta$ and the central angle $\alpha$ are supplementary:
$$\theta = 180^\circ - \alpha = 180^\circ - 110^\circ = 70^\circ$$
Final Answer:
$$\mathbf{\text{Angle between tangents} = 70^\circ}$$
Exercise 10.1 - Q1(1(vii)) Tangents & Radial Orthogonality: In the figure, a chord makes an angle of $45^\circ$ with the tangent line at the point of contact. Find the angle subtended by the chord in the alternate segment.
Step 1: Apply the Alternate Segment Theorem (Theorem 10.6):
The measure of the angle between a tangent and a chord drawn through the point of contact is equal to the angle in the alternate segment.
$$\theta = 45^\circ$$
Final Answer:
$$\mathbf{\text{Angle in alternate segment} = 45^\circ}$$
Exercise 10.1 - Q1(1(viii)) Tangents & Radial Orthogonality: In the figure, a secant line and tangent line intersect forming angles of $56^\circ$ and $60^\circ$. Find the unknown angles $x, y,$ and $z$.
Step 1: Compute $z$ using the triangle angle sum:
$$z = 180^\circ - 56^\circ - 60^\circ = 64^\circ$$
Step 2: Apply the Alternate Segment Theorem:
$$x = 64^\circ, \quad y = 56^\circ$$
Final Answer:
$$\mathbf{x = 64^\circ, \quad y = 56^\circ, \quad z = 60^\circ}$$
Exercise 10.1 - Q2 Tangents and Central Angle: In the figure, $AB$ and $AC$ are two tangents to a circle with centre $O$ at points $B$ and $C$ respectively. If $\angle BOC = 120^\circ$, find $\angle BAC$.
Step 1: Identify right angles at tangent contact points:
By Theorem 10.2, $OB \perp AB \implies \angle OBA = 90^\circ$ and $OC \perp AC \implies \angle OCA = 90^\circ$.
Step 2: Sum of angles in quadrilateral $OBAC$:
$$\angle BAC + \angle BOC + \angle OBA + \angle OCA = 360^\circ$$
$$\angle BAC + 120^\circ + 90^\circ + 90^\circ = 360^\circ \implies \angle BAC = 360^\circ - 300^\circ = 60^\circ$$
Final Answer:
$$\mathbf{\angle BAC = 60^\circ}$$
Exercise 10.1 - Q3 Tangent Chord Angles: In the figure, $AB$ and $AC$ are tangents from $A$ to the circle. If $\angle BAC = 50^\circ$, find the measure of $\angle OBC$.
Step 1: Find central angle $\angle BOC$:
$$\angle BOC = 180^\circ - \angle BAC = 180^\circ - 50^\circ = 130^\circ$$
Step 2: Use isosceles triangle $\triangle OBC$ ($OB = OC = r$):
$$\angle OBC = \frac{180^\circ - 130^\circ}{2} = \frac{50^\circ}{2} = 25^\circ$$
Final Answer:
$$\mathbf{\angle OBC = 25^\circ}$$
Exercise 10.1 - Q4 Alternate Segment Theorem Application: In the figure, line $CD$ is tangent to the circle at point $B$. If $\angle CBD = 50^\circ$, find the measure of $\angle CAB$ where $A$ lies on the alternate segment.
Step 1: Apply Theorem 10.6 (Alternate Segment Theorem):
The angle between tangent $CD$ and chord $AB$ at point $B$ is equal to the angle subtended by the chord in the alternate segment:
$$\angle CAB = \angle CBD = 50^\circ$$
Final Answer:
$$\mathbf{\angle CAB = 50^\circ}$$
Exercise 10.1 - Q5 Three Pairwise Touching Circles: Three circles touch each other externally. The distance between their centres are $8\text{ cm}$, $12\text{ cm}$, and $16\text{ cm}$. Find the radii of the circles.
Step 1: Formulate the system of equations using Theorem 10.4:
Let the radii of the three circles be $r_1, r_2, r_3$.
- $r_1 + r_2 = 8$ ...(1)
- $r_2 + r_3 = 12$ ...(2)
- $r_3 + r_1 = 16$ ...(3)
Step 2: Add all three equations:
$$2(r_1 + r_2 + r_3) = 8 + 12 + 16 = 36 \implies r_1 + r_2 + r_3 = 18\text{ cm}$$
Step 3: Solve for each individual radius:
$$r_1 = 18 - (r_2 + r_3) = 18 - 12 = 6\text{ cm}$$
$$r_2 = 18 - (r_3 + r_1) = 18 - 16 = 2\text{ cm}$$
$$r_3 = 18 - (r_1 + r_2) = 18 - 8 = 10\text{ cm}$$
Final Answer:
$$\mathbf{r_1 = 6\text{ cm}, \quad r_2 = 2\text{ cm}, \quad r_3 = 10\text{ cm}}$$
Exercise 10.1 - Q6 Distance Between Centres of Touching Circles: In the figure, $AD = 5\text{ cm}$, $CD = 4\text{ cm}$, and $BD = 4.5\text{ cm}$. Find the distance between the centres of the circles.
Step 1: Identify the radii of the two circles:
Radius of first circle $r_1 = AD = 5\text{ cm}$. Radius of second circle $r_2 = BD = 4.5\text{ cm}$.
Step 2: Calculate the distance between centres:
Since the circles touch externally at point $D$ on the line segment joining their centres:
$$\text{Distance between centres} = r_1 + r_2 = 5\text{ cm} + 4.5\text{ cm} = 9.5\text{ cm}$$
Final Answer:
$$\mathbf{\text{Distance} = 9.5\text{ cm}}$$
Exercise 10.1 - Q7 Midpoint & Right Angle Tangency: In the adjoining figure, $\angle PEF = 20^\circ$. $H$ is the midpoint of $EF$. Find $\angle PGH$.
Step 1: Use the right triangle geometry:
In $\triangle PEH$, $PH \perp EF$ because $H$ is the midpoint of chord $EF$ (Theorem 9.2). Therefore, $\angle PHE = 90^\circ$.
$$\angle EPH = 90^\circ - 20^\circ = 70^\circ$$
Step 2: Determine $\angle PGH$:
By symmetry and cyclic segment properties, $\angle PGH = \angle EPH = 70^\circ$.
Final Answer:
$$\mathbf{\angle PGH = 70^\circ}$$
Exercise 10.1 - Q8 Isosceles Triangle & Tangent Angles: In the figure, in triangle $ABC$, $AB = AC$. Find the values of angles $x$ and $y$.
Step 1: Use the Alternate Segment Theorem:
The angle between the tangent and chord equals the angle in the alternate segment.
$$x = 60^\circ$$
Step 2: Use isosceles triangle property ($AB = AC$):
$$\angle ABC = \angle ACB = 60^\circ \implies y = 180^\circ - (60^\circ + 60^\circ) = 60^\circ$$
Final Answer:
$$\mathbf{x = 60^\circ, \quad y = 60^\circ}$$
Exercise 10.2 • Angles in Segments, Alternate Segment Theorem & Cyclic Polygons
Exhaustive solutions for all 14 questions & subparts (Pages 226–229)
Exercise 10.2 - Q1 Inscribed & Central Angles: In the adjoining figure, $\angle P = 30^\circ$. Find the values of $\angle Q$ and $\angle ROS$.
Step 1: Identify Inscribed Angles Subtended by the Same Arc $RS$:
Both $\angle P$ and $\angle Q$ are inscribed angles subtended by the same chord / arc $RS$ in the same segment of the circle.
By Theorem 10.8 (Angles in the same segment of a circle are equal):
$$\angle Q = \angle P = 30^\circ$$
Step 2: Relate Central Angle $\angle ROS$ to Inscribed Angle $\angle P$:
By Theorem 10.7, the measure of a central angle of an arc of a circle is double the angle subtended by the corresponding major arc:
$$\angle ROS = 2 \times \angle P = 2 \times 30^\circ = 60^\circ$$
Final Answer:
$$\mathbf{\angle Q = 30^\circ}, \quad \mathbf{\angle ROS = 60^\circ}$$
Exercise 10.2 - Q2 Radii & Inscribed Angle: In the adjoining figure, $\angle BAO = 30^\circ$ where $O$ is the centre of the circle and $AB$ is a chord. Find $\angle ABO$, $\angle AOB$, and $\angle ACB$ (where $C$ lies on the circle).
Step 1: Use Isosceles Triangle Properties in $\triangle OAB$:
Since $OA$ and $OB$ are radii of the same circle, $OA = OB = r$.
Therefore, $\triangle OAB$ is an isosceles triangle with base angles equal:
$$\angle ABO = \angle BAO = 30^\circ$$
Step 2: Find Central Angle $\angle AOB$:
The sum of interior angles in $\triangle OAB$ is $180^\circ$:
$$\angle AOB = 180^\circ - (\angle BAO + \angle ABO) = 180^\circ - (30^\circ + 30^\circ) = 180^\circ - 60^\circ = 120^\circ$$
Step 3: Find Inscribed Angle $\angle ACB$:
By Theorem 10.7, the inscribed angle subtended by chord $AB$ on the major arc is half of the central angle:
$$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 120^\circ = 60^\circ$$
Final Answer:
$$\mathbf{\angle ABO = 30^\circ}, \quad \mathbf{\angle AOB = 120^\circ}, \quad \mathbf{\angle ACB = 60^\circ}$$
Exercise 10.2 - Q3 Cyclic Circle Angles: In the adjoining figure, $\angle BAC = 40^\circ$ and $\angle ACD = 60^\circ$. Points $A, B, C, D$ lie on the circumference of the circle. Find $\angle ABC$, $\angle BCA$, and $\angle CAD$.
Step 1: Identify Angles in the Same Segment:
By Theorem 10.8, angles subtended by chord $CD$ in the same segment are equal:
$$\angle CAD = \angle CBD \quad \text{and} \quad \angle BCA = \angle BDA$$
Also, chord $AD$ subtends $\angle ACD = 60^\circ \implies \angle ABD = 60^\circ$.
Step 2: Compute $\angle ABC$ and $\angle BCA$:
Using triangle and cyclic quadrilateral relations:
$$\angle BCA = \angle BDA = 60^\circ$$
In $\triangle ABC$:
$$\angle ABC = 180^\circ - (\angle BAC + \angle BCA) = 180^\circ - (40^\circ + 60^\circ) = 180^\circ - 100^\circ = 80^\circ$$
$$\angle CAD = 60^\circ$$
Final Answer:
$$\mathbf{\angle ABC = 80^\circ}, \quad \mathbf{\angle BCA = 60^\circ}, \quad \mathbf{\angle CAD = 60^\circ}$$
Exercise 10.2 - Q4 Equilateral Central Triangle: In the figure, $\triangle AOB$ is an equilateral triangle where $O$ is the centre of the circle and $A, B$ lie on the circle. Point $C$ lies on the major arc. Find the measures of $\angle AOB$ and $\angle ACB$.
Step 1: Measure of Central Angle $\angle AOB$:
Since $\triangle AOB$ is an equilateral triangle, all its interior angles are equal to $60^\circ$:
$$\angle AOB = 60^\circ$$
Step 2: Calculate Inscribed Angle $\angle ACB$:
By Theorem 10.7, the inscribed angle subtended by chord $AB$ at point $C$ on the circumference is half of the central angle subtended by the same arc:
$$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^\circ = 30^\circ$$
Final Answer:
$$\mathbf{\angle AOB = 60^\circ}, \quad \mathbf{\angle ACB = 30^\circ}$$
Exercise 10.2 - Q5 Parallel Chords & Intersecting Diagonals: In the figure, chord $AB$ is parallel to chord $CD$ in a circle, and chords $AD$ and $BC$ intersect at $E$. If $\angle AEB = 70^\circ$, find the values of $\angle B$, $\angle C$, and $\angle D$.
Step 1: Symmetry and Alternate Angles:
Since $AB \parallel CD$, the arc lengths $AC$ and $BD$ between parallel chords are equal. This makes $ABCD$ an isosceles trapezium inscribed in the circle, with diagonals $AD = BC$.
Step 2: Angle relations at intersection $E$:
In $\triangle ABE$, $AE = BE \implies \angle EAB = \angle EBA$.
$$\angle EAB + \angle EBA + \angle AEB = 180^\circ$$
$$2\angle B + 70^\circ = 180^\circ \implies 2\angle B = 110^\circ \implies \angle B = 55^\circ$$
Step 3: Compute $\angle C$ and $\angle D$:
By alternate interior angles between $AB \parallel CD$:
$$\angle C = \angle B = 55^\circ, \quad \angle D = \angle A = 55^\circ$$
Final Answer:
$$\mathbf{\angle AEB = 70^\circ}, \quad \mathbf{\angle B = 55^\circ}, \quad \mathbf{\angle C = 55^\circ}, \quad \mathbf{\angle D = 55^\circ}$$
Exercise 10.2 - Q6 Tangent, Diameter & Alternate Segment: In the figure, $AB = 6\text{ cm}$ is the diameter of a circle and $CD$ is a tangent line touching the circle at $B$. Point $E$ lies on the circle such that $\angle EAB = x = 30^\circ$. Find: (i) Radius of the circle (ii) $\angle ABE$ (iii) $\angle ABC$ (iv) $\angle EBD$.
(i) Find the radius of the circle:
$$\text{Radius } r = \frac{\text{Diameter } AB}{2} = \frac{6\text{ cm}}{2} = 3\text{ cm}$$
(ii) Find $\angle ABE$:
By Theorem 10.9, the angle inscribed in a semicircle is a right angle:
$$\angle AEB = 90^\circ$$
In right-angled triangle $\triangle ABE$:
$$\angle ABE = 180^\circ - (\angle AEB + \angle EAB) = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$$
(iii) Find $\angle ABC$:
By Theorem 10.2, the tangent line $CD$ is perpendicular to diameter $AB$ at the point of contact $B$:
$$\angle ABC = 90^\circ$$
(iv) Find $\angle EBD$:
By the Alternate Segment Theorem (Theorem 10.6), the angle between tangent $BD$ and chord $BE$ equals the angle in the alternate segment ($\angle EAB$):
$$\angle EBD = \angle EAB = 30^\circ$$
(Alternatively: $\angle EBD = 90^\circ - \angle ABE = 90^\circ - 60^\circ = 30^\circ$)
Final Answer:
$$\mathbf{(i)\ r = 3\text{ cm}}, \quad \mathbf{(ii)\ \angle ABE = 60^\circ}, \quad \mathbf{(iii)\ \angle ABC = 90^\circ}, \quad \mathbf{(iv)\ \angle EBD = 30^\circ}$$
Exercise 10.2 - Q7 Geometric Similarity & Square Geometry: In the figure, $ABCD$ is a square having side length $10\text{ cm}$. Triangle $BOP$ is enlarged to form triangle $BCQ$. Find: (i) Scale factor of enlargement if $OP = 2\text{ cm}$ (ii) Length of $CQ$ (iii) $\angle BQC$ and $\angle BPO$ (iv) $AO$ (v) $AP$.
(i) Find the Scale Factor $k$:
Side of square $BC = 10\text{ cm}$. Corresponding side in $\triangle BOP$ is $BO$. In right triangle setup, $BO = 2\text{ cm} \implies$ Scale factor:
$$k = \frac{BC}{BO} = \frac{10}{2} = 5$$
(ii) Find Length of $CQ$:
$$CQ = k \times OP = 5 \times 2\text{ cm} = 10\text{ cm}$$
(iii) Find $\angle BQC$ and $\angle BPO$:
Since $\triangle BCQ \sim \triangle BOP$ and $BC \perp CQ$:
$$\angle BQC = 90^\circ \implies \angle BPO = 90^\circ$$
(iv) Find Length $AO$:
$$AO = AB - BO = 10\text{ cm} - 2\text{ cm} = 8\text{ cm}$$
(v) Find Length $AP$:
Using Pythagorean theorem in $\triangle AOP$ where hypotenuse is $AO = 8\text{ cm}$ and $OP = 2\text{ cm}$:
$$AP = \sqrt{AO^2 - OP^2} = \sqrt{8^2 - 2^2} = \sqrt{64 - 4} = \sqrt{60} \approx 7.75\text{ cm} \quad (\text{or integer configuration } 6\text{ cm})$$
Final Answer:
$$\mathbf{(i)\ k = 5}, \quad \mathbf{(ii)\ CQ = 10\text{ cm}}, \quad \mathbf{(iii)\ \angle BQC = \angle BPO = 90^\circ}, \quad \mathbf{(iv)\ AO = 8\text{ cm}}, \quad \mathbf{(v)\ AP = \sqrt{60}\text{ cm} \approx 7.75\text{ cm}}$$
Exercise 10.2 - Q8 Angles in the Same Segment & Intersecting Chords: In the adjoining figure, $\angle QPR = 60^\circ$ and chords $QS, PR$ intersect at $T$. Find: (i) $\angle QSR$ (ii) $\angle PRS$ (iii) $\angle PQS$ (iv) $\angle QTR$ (v) Why is $\angle QTR$ an acute angle?
(i) Find $\angle QSR$:
By Theorem 10.8 (Angles in the same segment are equal):
Both $\angle QPR$ and $\angle QSR$ are subtended by the same chord $QR$.
$$\angle QSR = \angle QPR = 60^\circ$$
(ii) & (iii) Find $\angle PRS$ and $\angle PQS$:
By the same theorem, both subtend chord $PS$:
$$\angle PQS = \angle PRS = 60^\circ$$
(iv) Find $\angle QTR$:
In $\triangle TQR$, the sum of angles is $180^\circ$:
$$\angle QTR = 180^\circ - (\angle TQR + \angle TRQ) = 180^\circ - (60^\circ + 60^\circ) = 60^\circ$$
(v) Why $\angle QTR$ is an acute angle:
An angle is acute if its measure is strictly less than $90^\circ$. Here $\angle QTR = 60^\circ < 90^\circ$. Geometrically, the angle formed by chords intersecting inside a circle is the average of intercepted arcs, which in this equilateral configuration is $60^\circ < 90^\circ$.
Final Answer:
$$\mathbf{(i)\ \angle QSR = 60^\circ}, \quad \mathbf{(ii)\ \angle PRS = 60^\circ}, \quad \mathbf{(iii)\ \angle PQS = 60^\circ}, \quad \mathbf{(iv)\ \angle QTR = 60^\circ}, \quad \mathbf{(v)\ \text{Acute because } 60^\circ < 90^\circ}$$
Exercise 10.2 - Q9 Central Angle & Cyclic Quadrilateral: In the given figure, $P$ is the centre of the circle and $\angle BPD = 130^\circ$. Points $A, B, C, D$ lie on the circle forming cyclic quadrilateral $ABCD$. Find: (i) $\angle BAD$ (ii) $\angle BCD$ (iii) $\angle ABC + \angle ADC$.
(i) Find $\angle BAD$:
Central angle subtended by chord $BD$ is $\angle BPD = 130^\circ$.
By Theorem 10.7, the angle subtended by the arc at the circumference is half the central angle:
$$\angle BAD = \frac{1}{2} \angle BPD = \frac{1}{2} \times 130^\circ = 65^\circ$$
(ii) Find $\angle BCD$:
Since $ABCD$ is a cyclic quadrilateral, by Theorem 10.12, opposite angles are supplementary:
$$\angle BAD + \angle BCD = 180^\circ$$
$$\angle BCD = 180^\circ - \angle BAD = 180^\circ - 65^\circ = 115^\circ$$
(iii) Find $\angle ABC + \angle ADC$:
By Theorem 10.12, the other pair of opposite angles in cyclic quadrilateral $ABCD$ must also be supplementary:
$$\angle ABC + \angle ADC = 180^\circ$$
Final Answer:
$$\mathbf{(i)\ \angle BAD = 65^\circ}, \quad \mathbf{(ii)\ \angle BCD = 115^\circ}, \quad \mathbf{(iii)\ \angle ABC + \angle ADC = 180^\circ}$$
Exercise 10.2 - Q10 Cyclic Quadrilateral Angle Sums: In the given figure, $PQRS$ is a cyclic quadrilateral and $\angle QPS = 95^\circ$. If $PQ = PS$ and $\angle PQS = 40^\circ$, find: (i) $\angle QRS$ (ii) $\angle PQR$ (iii) $\angle PSR$.
(i) Find $\angle QRS$:
By Theorem 10.12, opposite angles of a cyclic quadrilateral are supplementary:
$$\angle QPS + \angle QRS = 180^\circ$$
$$\angle QRS = 180^\circ - 95^\circ = 85^\circ$$
(ii) Find $\angle PQR$:
In cyclic quadrilateral $PQRS$, opposite angles $\angle PQR + \angle PSR = 180^\circ$.
For isosceles cyclic setup:
$$\angle PQR = 180^\circ - 95^\circ = 85^\circ$$
(iii) Find $\angle PSR$:
$$\angle PSR = 180^\circ - \angle PQR = 180^\circ - 85^\circ = 95^\circ$$
Final Answer:
$$\mathbf{(i)\ \angle QRS = 85^\circ}, \quad \mathbf{(ii)\ \angle PQR = 85^\circ}, \quad \mathbf{(iii)\ \angle PSR = 95^\circ}$$
Exercise 10.2 - Q11 Cyclic Quadrilateral & Isosceles Chord Triangle: In the adjoining figure, $ABCD$ is a cyclic quadrilateral and $\angle ADC = 120^\circ$. (i) Find $\angle BAC$ (where $BC$ is diameter subtending $\angle BDC$). (ii) Find $\angle CAD$ if $AD = CD$.
(i) Find $\angle BAC$:
Opposite angles in cyclic quadrilateral $ABCD$ are supplementary:
$$\angle ABC + \angle ADC = 180^\circ \implies \angle B = 180^\circ - 120^\circ = 60^\circ$$
By Theorem 10.8, angles in the same segment subtended by chord $BC$ give:
$$\angle BAC = 180^\circ - 120^\circ = 60^\circ$$
(ii) Find $\angle CAD$ if $AD = CD$:
In $\triangle ACD$, since $AD = CD$, it is an isosceles triangle with $\angle DAC = \angle DCA$.
The angle sum in $\triangle ACD$ is $180^\circ$:
$$\angle ADC + \angle CAD + \angle ACD = 180^\circ$$
$$120^\circ + 2\angle CAD = 180^\circ \implies 2\angle CAD = 60^\circ \implies \angle CAD = 30^\circ$$
Final Answer:
$$\mathbf{(i)\ \angle BAC = 60^\circ}, \quad \mathbf{(ii)\ \angle CAD = 30^\circ}$$
Exercise 10.2 - Q12 Cyclic Quadrilateral Exterior Angles: In the figure, cyclic quadrilateral $ABCD$ has side $AB$ extended to $E$ such that exterior angle $\angle ABE = 95^\circ$, and $\angle ADE = 65^\circ$. Find: (i) $\angle ECD$ (ii) $\angle CDE$ (iii) $\angle CED$.
(i) Find $\angle ECD$:
By the Exterior Angle Property of a Cyclic Quadrilateral, the exterior angle at any vertex is equal to the interior opposite angle:
$$\angle ECD = \angle ABE = 95^\circ$$
(ii) Find $\angle CDE$:
Given that line $DE$ forms angle $\angle ADE = 65^\circ \implies \angle CDE = 65^\circ$.
(iii) Find $\angle CED$:
In $\triangle CDE$, the sum of interior angles is $180^\circ$:
$$\angle CED + \angle ECD + \angle CDE = 180^\circ$$
$$\angle CED + 95^\circ + 65^\circ = 180^\circ$$
$$\angle CED + 160^\circ = 180^\circ \implies \angle CED = 180^\circ - 160^\circ = 20^\circ$$
Final Answer:
$$\mathbf{(i)\ \angle ECD = 95^\circ}, \quad \mathbf{(ii)\ \angle CDE = 65^\circ}, \quad \mathbf{(iii)\ \angle CED = 20^\circ}$$
Exercise 10.2 - Q13 Tangent Application (Hanging Wall Clock): A circular wall clock of radius $1\text{ ft}$ is hung by a $4\text{ feet}$ long string with a nail at $P$. The string forms two symmetric tangents to the circular clock. (i) Find the distance between the centre of the clock and the nail. (ii) How far above the clock is the nail?
Step 1: Identify Geometry of Hanging Clock:
Let $O$ be the center of the circular clock of radius $r = 1\text{ ft}$.
The string of total length $4\text{ ft}$ is tied around the nail at $P$ and touches the clock tangentially at two points $A$ and $B$.
By symmetry, the two tangent segments from nail $P$ to contact points $A$ and $B$ are each:
$$\text{Tangent length } t = PA = PB = \frac{4\text{ ft}}{2} = 2\text{ ft}$$
Step 2: (i) Distance between Centre $O$ and Nail $P$:
By Theorem 10.2, the radius $OA \perp PA$, forming right-angled triangle $\triangle OAP$ with hypotenuse $OP$:
$$OP^2 = OA^2 + PA^2 = 1^2 + 2^2 = 1 + 4 = 5$$
$$OP = \sqrt{5}\text{ ft} \approx 2.236\text{ ft}$$
Step 3: (ii) How far above the clock is the nail:
The highest point on the circular clock is at distance $r = 1\text{ ft}$ from center $O$ along the line $OP$.
$$\text{Distance above clock} = OP - r = \sqrt{5} - 1 \approx 2.236 - 1 = 1.236\text{ ft}$$
Final Answer:
$$\mathbf{(i)\ \text{Distance from centre to nail } OP = \sqrt{5}\text{ ft} \approx 2.24\text{ ft}}$$
$$\mathbf{(ii)\ \text{Distance of nail above clock } = \sqrt{5} - 1\text{ ft} \approx 1.24\text{ ft}}$$
Exercise 10.2 - Q14 Tangent Application (Wheel on Flat Ground): A wooden wheel moving on the ground is $99\text{ cm}$ away from a point $A$ on the ground (measured from the contact point $T$). Find the diameter of the wheel if the distance between the centre of the wheel $O$ and point $A$ is $101\text{ cm}$. How much distance does the wheel cover in one complete round? (Take $\pi \approx \frac{22}{7}$)
Step 1: Set up the Right Triangle $\triangle OTA$:
The ground is a tangent to the circular wheel at contact point $T$.
By Theorem 10.2, the radius $OT \perp AT$, forming right triangle $\triangle OTA$ where:
- Hypotenuse $OA = 101\text{ cm}$
- Tangent leg $AT = 99\text{ cm}$
- Radius leg $r = OT$
Step 2: Calculate Radius $r$ using Pythagorean Theorem:
$$r^2 = OA^2 - AT^2 = 101^2 - 99^2$$
Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$:
$$r^2 = (101 - 99)(101 + 99) = 2 \times 200 = 400$$
$$r = \sqrt{400} = 20\text{ cm}$$
Step 3: Calculate Diameter $d$:
$$d = 2r = 2 \times 20\text{ cm} = 40\text{ cm}$$
Step 4: Calculate Distance covered in one round (Circumference $C$):
$$C = 2\pi r = \pi d = \frac{22}{7} \times 40 = \frac{880}{7} \approx 125.71\text{ cm}$$
Final Answer:
$$\mathbf{\text{Diameter of wheel } d = 40\text{ cm}}$$
$$\mathbf{\text{Distance covered in one round } C \approx 125.71\text{ cm}}$$
Miscellaneous Exercise 10 • Comprehensive Chapter Review
Exhaustive solutions for all 23 questions (20 MCQs + 3 Analytical Problems) (Pages 230–231)
Miscellaneous Exercise 10 - Q1(i) Objective: A tangent line touches the circle at ______ point(s).
Explanation: By definition, a tangent to a circle is a straight line that intersects or touches the circle at exactly one single point, known as the point of contact.
Miscellaneous Exercise 10 - Q1(ii) Objective: A tangent line is ______ to the radial segment at the point of contact.
Explanation: By Theorem 10.2, the tangent at any point of a circle is perpendicular to the radial segment through the point of contact ($m\angle = 90^\circ$).
Miscellaneous Exercise 10 - Q1(iii) Objective: How many tangents can be drawn to a circle from a point outside the circle?
Explanation: From any external point outside a circle, exactly two distinct tangents can be drawn to the circle (Theorem 10.3).
Miscellaneous Exercise 10 - Q1(iv) Objective: If two tangents are drawn at both ends of a diameter of a circle, they are:
Explanation: Since the diameter is a straight line segment and tangents at both ends are perpendicular to the diameter, two lines perpendicular to the same transversal are parallel ($90^\circ + 90^\circ = 180^\circ$).
Miscellaneous Exercise 10 - Q1(v) Objective: Given that the radius of a circle is $4\text{ cm}$. The distance between two parallel tangents drawn at the outer ends of a diameter is:
Explanation: The distance between two parallel tangents drawn at the endpoints of a diameter is equal to the length of the diameter: $d = 2r = 2 \times 4\text{ cm} = 8\text{ cm}$.
Miscellaneous Exercise 10 - Q1(vi) Objective: How many tangents can be drawn to a circle from a point lying on the circle?
Explanation: At any single point lying on the circumference of a circle, only one unique tangent line can be drawn.
Miscellaneous Exercise 10 - Q1(vii) Objective: Two tangents drawn from a point outside the circle are:
Explanation: By Theorem 10.3, the lengths of two tangents drawn from an external point to a circle are equal (congruent, $AB = AC$).
Miscellaneous Exercise 10 - Q1(viii) Objective: If two circles touch externally, the distance between their centres is equal to the sum of ______ of both circles.
Explanation: By Theorem 10.4, when two circles touch externally, the distance between their centres $d = r_1 + r_2$.
Miscellaneous Exercise 10 - Q1(ix) Objective: If two congruent circles touch externally, the distance between their centres is equal to the ______ of a circle.
Explanation: For congruent circles, $r_1 = r_2 = r$. The distance between centres is $d = r + r = 2r = \text{diameter}$.
Miscellaneous Exercise 10 - Q1(x) Objective: If two circles of radii $1.4\text{ cm}$ and $2.5\text{ cm}$ touch internally, the distance between their centres is equal to:
Explanation: By Theorem 10.5, the distance between centres of internally touching circles is $d = r_1 - r_2 = 2.5\text{ cm} - 1.4\text{ cm} = 1.1\text{ cm}$.
Miscellaneous Exercise 10 - Q1(xi) Objective: The angle subtended by an arc at the centre of a circle is called a(n) ______ angle.
Explanation: An angle whose vertex is at the center of a circle and whose arms are radii is called a central angle.
Miscellaneous Exercise 10 - Q1(xii) Objective: An angle inscribed in a semi-circle is a ______ angle.
Explanation: By Theorem 10.9, any angle inscribed in a semicircle is a right angle ($90^\circ$).
Miscellaneous Exercise 10 - Q1(xiii) Objective: If the central angle of a minor arc of a circle is $100^\circ$, the angle inscribed in the corresponding major arc is:
Explanation: By Theorem 10.7, the inscribed angle is half of the central angle subtended by the same arc: $\frac{100^\circ}{2} = 50^\circ$.
Miscellaneous Exercise 10 - Q1(xiv) Objective: The central angle of a minor arc of a circle is:
Explanation: By definition, a minor arc spans less than a semicircle, so its central angle is strictly less than $180^\circ$.
Miscellaneous Exercise 10 - Q1(xv) Objective: The central angle of a major arc of a circle is:
Explanation: A major arc spans more than half a circle, so its central angle is a reflex angle ($> 180^\circ$).
Miscellaneous Exercise 10 - Q1(xvi) Objective: All angles in the same segment of a circle are:
Explanation: By Theorem 10.8, any two or more angles inscribed in the same segment of a circle are congruent (equal in measure).
Miscellaneous Exercise 10 - Q1(xvii) Objective: If $ABCD$ is a cyclic quadrilateral and $\angle A = 60^\circ$, then $\angle C = $
Explanation: In a cyclic quadrilateral, opposite angles are supplementary (Theorem 10.12): $\angle A + \angle C = 180^\circ \implies \angle C = 180^\circ - 60^\circ = 120^\circ$.
Miscellaneous Exercise 10 - Q1(xviii) Objective: An exterior angle of a cyclic quadrilateral is equal to the ______ interior angle.
Explanation: The exterior angle of any cyclic quadrilateral is equal in measure to its interior opposite angle.
Miscellaneous Exercise 10 - Q1(xix) Objective: The inscribed angle of a quadrant of a circle is:
Explanation: A quadrant arc has central angle $90^\circ$. The remaining major arc has reflex central angle $360^\circ - 90^\circ = 270^\circ$. The inscribed angle in the minor segment is $\frac{270^\circ}{2} = 135^\circ$.
Miscellaneous Exercise 10 - Q1(xx) Objective: In a circle with central angle $\angle AOB = 94^\circ$, the inscribed angle $\theta = \angle ACB$ is equal to:
Explanation: By Theorem 10.7, the inscribed angle is half of the central angle: $\theta = \frac{94^\circ}{2} = 47^\circ$.
Miscellaneous Exercise 10 - Q2 Chord Subtending Right Central Angle: In a circle whose diameter is $12\text{ cm}$, there is a central angle whose measure is $90^\circ$. A chord joins the endpoints of the arc cut off by the angle. Find the length of the chord.
Step 1: Determine the Radius of the Circle:
$$\text{Radius } r = \frac{\text{Diameter}}{2} = \frac{12\text{ cm}}{2} = 6\text{ cm}$$
Step 2: Set up Right-Angled Triangle $\triangle AOB$:
Let $O$ be the centre and $A, B$ be endpoints of the chord. Since the central angle $\angle AOB = 90^\circ$, $\triangle AOB$ is a right-angled isosceles triangle with legs $OA = OB = r = 6\text{ cm}$.
Step 3: Calculate Chord Length $AB$ via Pythagorean Theorem:
$$AB^2 = OA^2 + OB^2 = 6^2 + 6^2 = 36 + 36 = 72$$
$$AB = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\text{ cm} \approx 8.49\text{ cm}$$
Final Answer:
$$\mathbf{\text{Length of the chord } = 6\sqrt{2}\text{ cm} \approx 8.49\text{ cm}}$$
Miscellaneous Exercise 10 - Q3 Perpendicular Distance from Center to Parallel Chord: The diameter of a circle is $20\text{ cm}$ long and a chord parallel to it is $12\text{ cm}$ long. Find the distance between the chord and the center of the circle.
Step 1: Identify Radius and Half-Chord Length:
$$\text{Radius } r = \frac{\text{Diameter}}{2} = \frac{20\text{ cm}}{2} = 10\text{ cm}$$
Let the chord length be $AB = 12\text{ cm}$.
The perpendicular drawn from the center $O$ to chord $AB$ bisects the chord at point $M$:
$$AM = MB = \frac{AB}{2} = \frac{12\text{ cm}}{2} = 6\text{ cm}$$
Step 2: Calculate Perpendicular Distance $OM$:
In right-angled triangle $\triangle OMA$ (where $OA = r = 10\text{ cm}$):
$$OM^2 + AM^2 = OA^2$$
$$OM^2 + 6^2 = 10^2 \implies OM^2 + 36 = 100$$
$$OM^2 = 100 - 36 = 64 \implies OM = \sqrt{64} = 8\text{ cm}$$
Final Answer:
$$\mathbf{\text{Distance between the chord and the center } = 8\text{ cm}}$$
Miscellaneous Exercise 10 - Q4 Ptolemy's Theorem for Cyclic Quadrilaterals: In a cyclic quadrilateral $ABCD$ with sides $a = AB = 4\text{ cm}$, $b = BC = 5\text{ cm}$, $c = CD = 6\text{ cm}$, and $d = DA = 3\text{ cm}$, and diagonal $p = AC = 8\text{ cm}$. Find diagonal $q = BD$ using Ptolemy's Theorem.
Step 1: State Ptolemy's Theorem for Cyclic Quadrilaterals:
For any quadrilateral $ABCD$ inscribed in a circle, the product of its diagonals equals the sum of the products of its opposite sides:
$$p \cdot q = (a \cdot c) + (b \cdot d)$$
where $p = AC$, $q = BD$, $a = AB$, $b = BC$, $c = CD$, and $d = DA$.
Step 2: Substitute Given Numerical Values:
$$a = 4\text{ cm}, \quad b = 5\text{ cm}, \quad c = 6\text{ cm}, \quad d = 3\text{ cm}, \quad p = 8\text{ cm}$$
$$8 \times q = (4 \times 6) + (5 \times 3)$$
$$8q = 24 + 15 = 39$$
Step 3: Solve for Diagonal $q$:
$$q = \frac{39}{8} = 4.875\text{ cm}$$
Final Answer:
$$\mathbf{q = BD = \frac{39}{8}\text{ cm} = 4.875\text{ cm}}$$
❓ Frequently Asked Questions — Class 10 Mathematics Chapter 10
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