Class 10 Mathematics - Ch 10: Tangent to a Circle
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 10: Tangent to a Circle
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
🎯 Question Types & Curriculum Breakdown
Total Question Pool
100%
144 Questions
Combined Active Syllabus
Multiple Choice (MCQs)
39%
56 MCQs
Available
Short Questions
37%
53 Questions
Available
Long / Theory Questions
8%
11 Questions
Available
True / False
7%
10 Questions
Available
Fill in the Blanks
7%
10 Questions
Available
Match Columns
3%
4 Questions
Available
📊 Question Pool Structure
144 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 10: Tangent to a Circle, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
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📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
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 $x$.
✓ Correct Answer: 5 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Apply the Radius-Tangent Perpendicularity Theorem (Theorem 10.2):</strong></p>
<p>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 $b = 12\text{ cm}$.</p>
<p><strong>Step 2: Apply the Pythagorean Theorem:</strong></p>
<p>$$x^2 + 12^2 = 13^2$$</p>
<p>$$x^2 + 144 = 169 \implies x^2 = 169 - 144 = 25$$</p>
<p>$$x = \sqrt{25} = 5\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{x = 5\text{ cm}}$$</p>
<p>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 $b = 12\text{ cm}$.</p>
<p><strong>Step 2: Apply the Pythagorean Theorem:</strong></p>
<p>$$x^2 + 12^2 = 13^2$$</p>
<p>$$x^2 + 144 = 169 \implies x^2 = 169 - 144 = 25$$</p>
<p>$$x = \sqrt{25} = 5\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{x = 5\text{ cm}}$$</p>
Sample Question 2
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
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.
✓ Correct Answer: 2.9 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Set up the right triangle relations:</strong></p>
<p>Radius $r = 2\text{ cm}$, Tangent length $t = 2.1\text{ cm}$. The distance from the center to the external point is the hypotenuse $d$.</p>
<p><strong>Step 2: Calculate hypotenuse using Pythagoras Theorem:</strong></p>
<p>$$d = \sqrt{r^2 + t^2} = \sqrt{2^2 + (2.1)^2} = \sqrt{4 + 4.41} = \sqrt{8.41} = 2.9\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 2.9\text{ cm}}$$</p>
<p>Radius $r = 2\text{ cm}$, Tangent length $t = 2.1\text{ cm}$. The distance from the center to the external point is the hypotenuse $d$.</p>
<p><strong>Step 2: Calculate hypotenuse using Pythagoras Theorem:</strong></p>
<p>$$d = \sqrt{r^2 + t^2} = \sqrt{2^2 + (2.1)^2} = \sqrt{4 + 4.41} = \sqrt{8.41} = 2.9\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 2.9\text{ cm}}$$</p>
Sample Question 3
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
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 $x$.
✓ Correct Answer: 15 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Set up the Pythagorean equation:</strong></p>
<p>Hypotenuse $OP = 17\text{ cm}$, Radius $r = 8\text{ cm}$, Tangent $x$.</p>
<p>$$x = \sqrt{OP^2 - r^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{x = 15\text{ cm}}$$</p>
<p>Hypotenuse $OP = 17\text{ cm}$, Radius $r = 8\text{ cm}$, Tangent $x$.</p>
<p>$$x = \sqrt{OP^2 - r^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{x = 15\text{ cm}}$$</p>
Sample Question 4
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
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.
✓ Correct Answer: 6√3 ≈ 10.39 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Apply the Pythagorean Theorem:</strong></p>
<p>$$d = \sqrt{r^2 + t^2} = \sqrt{6^2 + (6\sqrt{2})^2} = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}}$$</p>
<p>$$d = \sqrt{r^2 + t^2} = \sqrt{6^2 + (6\sqrt{2})^2} = \sqrt{36 + 72} = \sqrt{108} = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 6\sqrt{3}\text{ cm} \approx 10.39\text{ cm}}$$</p>
Sample Question 5
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
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.
✓ Correct Answer: 130°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Apply the Quadrilateral Angle Sum Property:</strong></p>
<p>In the quadrilateral formed by the two radii, the two tangents, and the center, the angles at the points of contact are both $90^\circ$.</p>
<p>$$\text{Angle between tangents} + \text{Central angle} = 180^\circ$$</p>
<p>$$\text{Central angle} = 180^\circ - 50^\circ = 130^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Central Angle} = 130^\circ}$$</p>
<p>In the quadrilateral formed by the two radii, the two tangents, and the center, the angles at the points of contact are both $90^\circ$.</p>
<p>$$\text{Angle between tangents} + \text{Central angle} = 180^\circ$$</p>
<p>$$\text{Central angle} = 180^\circ - 50^\circ = 130^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{\text{Central Angle} = 130^\circ}$$</p>
Sample Question 6
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
The central angle between two radial segments drawn to the points of contact of two tangents is $120^\circ$. Find the angle between the two tangents.
✓ Correct Answer: 60°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Calculate the supplementary angle:</strong></p>
<p>$$\text{Angle between tangents} = 180^\circ - \text{Central angle} = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{60^\circ}$$</p>
<p>$$\text{Angle between tangents} = 180^\circ - \text{Central angle} = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{60^\circ}$$</p>
Sample Question 7
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
Two circles of radii $15\text{ cm}$ and $8\text{ cm}$ touch externally. Find the distance between their centres.
✓ Correct Answer: 23 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Apply Theorem 10.4 (Externally Touching Circles):</strong></p>
<p>When two circles touch externally, the distance between their centres equals the sum of their radii:</p>
<p>$$d = r_1 + r_2 = 15\text{ cm} + 8\text{ cm} = 23\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 23\text{ cm}}$$</p>
<p>When two circles touch externally, the distance between their centres equals the sum of their radii:</p>
<p>$$d = r_1 + r_2 = 15\text{ cm} + 8\text{ cm} = 23\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 23\text{ cm}}$$</p>
Sample Question 8
Exercise 10.1 - Tangent & Touching Circles Geometry
MEDIUM • Short Question
Two circles of radii $12\text{ cm}$ and $5\text{ cm}$ touch internally. Find the distance between their centres.
✓ Correct Answer: 7 cm
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Step 1: Apply Theorem 10.5 (Internally Touching Circles):</strong></p>
<p>When two circles touch internally, the distance between their centres equals the difference of their radii:</p>
<p>$$d = r_1 - r_2 = 12\text{ cm} - 5\text{ cm} = 7\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 7\text{ cm}}$$</p>
<p>When two circles touch internally, the distance between their centres equals the difference of their radii:</p>
<p>$$d = r_1 - r_2 = 12\text{ cm} - 5\text{ cm} = 7\text{ cm}$$</p>
<p><strong>Final Answer:</strong></p>
<p>$$\mathbf{d = 7\text{ cm}}$$</p>
Sample Question 9
Exercise 10.1 - Tangents AB and AC with Central Angle 120°
MEDIUM • Short Question
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: (i) $\angle OBA$, (ii) $\angle OCB$, (iii) $\angle BAC$, (iv) $\angle ABC$.
✓ Correct Answer: ∠OBA = 90°, ∠OCB = 30°, ∠BAC = 60°, ∠ABC = 60°
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>(i) Find $\angle OBA$:</strong></p>
<p>By Theorem 10.2, the radial segment $OB$ is perpendicular to tangent $AB$ at the point of contact $B$.</p>
<p>$$\angle OBA = 90^\circ$$</p>
<p><strong>(ii) Find $\angle OCB$:</strong></p>
<p>In $\triangle OBC$, $OB = OC = r$ (radii of the same circle). Thus, $\triangle OBC$ is isosceles with $\angle OBC = \angle OCB$.</p>
<p>$$\angle OCB = \frac{180^\circ - \angle BOC}{2} = \frac{180^\circ - 120^\circ}{2} = \frac{60^\circ}{2} = 30^\circ$$</p>
<p><strong>(iii) Find $\angle BAC$:</strong></p>
<p>In quadrilateral $ABOC$, the sum of angles is $360^\circ$. Since $\angle OBA = \angle OCA = 90^\circ$:</p>
<p>$$\angle BAC = 180^\circ - \angle BOC = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>(iv) Find $\angle ABC$:</strong></p>
<p>$$\angle ABC = \angle OBA - \angle OBC = 90^\circ - 30^\circ = 60^\circ$$</p>
<p><strong>Final Answers:</strong></p>
<p>$$\mathbf{\text{(i) } 90^\circ, \quad \text{(ii) } 30^\circ, \quad \text{(iii) } 60^\circ, \quad \text{(iv) } 60^\circ}$$</p>
<p>By Theorem 10.2, the radial segment $OB$ is perpendicular to tangent $AB$ at the point of contact $B$.</p>
<p>$$\angle OBA = 90^\circ$$</p>
<p><strong>(ii) Find $\angle OCB$:</strong></p>
<p>In $\triangle OBC$, $OB = OC = r$ (radii of the same circle). Thus, $\triangle OBC$ is isosceles with $\angle OBC = \angle OCB$.</p>
<p>$$\angle OCB = \frac{180^\circ - \angle BOC}{2} = \frac{180^\circ - 120^\circ}{2} = \frac{60^\circ}{2} = 30^\circ$$</p>
<p><strong>(iii) Find $\angle BAC$:</strong></p>
<p>In quadrilateral $ABOC$, the sum of angles is $360^\circ$. Since $\angle OBA = \angle OCA = 90^\circ$:</p>
<p>$$\angle BAC = 180^\circ - \angle BOC = 180^\circ - 120^\circ = 60^\circ$$</p>
<p><strong>(iv) Find $\angle ABC$:</strong></p>
<p>$$\angle ABC = \angle OBA - \angle OBC = 90^\circ - 30^\circ = 60^\circ$$</p>
<p><strong>Final Answers:</strong></p>
<p>$$\mathbf{\text{(i) } 90^\circ, \quad \text{(ii) } 30^\circ, \quad \text{(iii) } 60^\circ, \quad \text{(iv) } 60^\circ}$$</p>
Sample Question 10
Exercise 10.1 - Equal Chords of Concentric Circles Tangent to Inner Circle
MEDIUM • Short Question
$O$ is the centre of two concentric circles. $AB$ and $CD$ are chords of the outer circle that are tangent to the inner circle. Prove that $AB = CD$.
✓ Correct Answer: AB = CD (Proved by Theorem 9.5 & 10.2)
📖 Step-by-Step Solution & Conceptual Rationale:
<p><strong>Given:</strong> Two concentric circles with common centre $O$. Chords $AB$ and $CD$ of the outer circle are tangents to the inner circle of radius $r$.</p>
<p><strong>To Prove:</strong> $\text{Chord } AB = \text{Chord } CD$.</p>
<p><strong>Proof:</strong></p>
<ol>
<li>Let $E$ and $F$ be the points of contact where $AB$ and $CD$ touch the inner circle.</li>
<li>By Theorem 10.2, the radial segments $OE \perp AB$ and $OF \perp CD$.</li>
<li>Since $E$ and $F$ lie on the inner circle, $OE = OF = r$ (radius of the inner circle).</li>
<li>Thus, the chords $AB$ and $CD$ of the outer circle are equidistant from the common centre $O$ ($OE = OF = r$).</li>
<li>By Theorem 9.5, chords of a circle equidistant from the centre are congruent.</li>
<li>Therefore, $AB = CD$.</li>
</ol>
<p><strong>Conclusion:</strong></p>
<p>$$\mathbf{AB = CD \quad \text{(Hence Proved)}}$$</p>
<p><strong>To Prove:</strong> $\text{Chord } AB = \text{Chord } CD$.</p>
<p><strong>Proof:</strong></p>
<ol>
<li>Let $E$ and $F$ be the points of contact where $AB$ and $CD$ touch the inner circle.</li>
<li>By Theorem 10.2, the radial segments $OE \perp AB$ and $OF \perp CD$.</li>
<li>Since $E$ and $F$ lie on the inner circle, $OE = OF = r$ (radius of the inner circle).</li>
<li>Thus, the chords $AB$ and $CD$ of the outer circle are equidistant from the common centre $O$ ($OE = OF = r$).</li>
<li>By Theorem 9.5, chords of a circle equidistant from the centre are congruent.</li>
<li>Therefore, $AB = CD$.</li>
</ol>
<p><strong>Conclusion:</strong></p>
<p>$$\mathbf{AB = CD \quad \text{(Hence Proved)}}$$</p>
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