Lengths of Tangents from an External Point for CBSE Class 10 Mathematics

How many tangents can you draw to a circle from a single point? If the point lies inside the circle, no line can ever touch without slicing through as a secant. If the point lies on the circumference, exactly one tangent can be drawn. But if you stand outside a circular boundary and draw straight lines touching the circle, you will find that you can draw exactly two tangents—and remarkably, both lines measure the exact same length.

In CBSE Class 10 Mathematics, Chapter 10 (Circles), Theorem 10.2 establishes this equality. This single theorem serves as the primary engine for solving virtually all circle riders, quadrilateral proofs, and geometric calculations in board examinations.

What You Will Learn

• Number of tangents drawn from points inside, on, and outside a circle
• Formal definition of the length of a tangent
• Statement and rigorous step-by-step geometric proof of Theorem 10.2 using RHS Congruence
• Three vital corollaries: Equal subtended center angles, angle bisector property, and supplementary angles
• Solved CBSE board examination problems (concentric circle chord bisectors)
• Presentation guidelines for full marks in Section C and D

1. Tangents from Different Points Relative to a Circle

    Case 1: Point Inside Circle       Case 2: Point On Circle         Case 3: Point Outside Circle
               O                               O                               O
              /                                |                              /              P (Inside)                        P (On)                        /             (0 Tangents possible;           (Exactly 1 Tangent)               P (Outside: Exactly 2 Tangents)
           all lines are secants)

1. Point $P$ inside the circle: Zero tangents can be drawn (every line passing through $P$ intersects the circle at two points).
2. Point $P$ on the circle: Exactly one tangent can be drawn.
3. Point $P$ outside the circle: Exactly two tangents can be drawn.

Definition of Length of Tangent:

The length of the tangent from an external point $P$ to a circle is the straight-line distance from the external point $P$ to the point of contact with the circle.

2. Theorem 10.2: Equal Tangent Lengths

Theorem Statement (CBSE Theorem 10.2)

The lengths of tangents drawn from an external point to a circle are equal.

                                      Q (Point of Contact 1)
                                     /|
                            Radius r/ |
                                   /  |
                    (Center) O ---+   | Tangent 1
                                   \  |
                            Radius r\ |
                                     \|
                                      R (Point of Contact 2)
                                                                                                                        P (External Point)

Given:

A circle with center $O$, an external point $P$, and two tangents $PQ$ and $PR$ touching the circle at points $Q$ and $R$ respectively.

To Prove:

$PQ = PR$

Construction:

Join $OP$, $OQ$, and $OR$.

Step-by-Step Geometric Proof:

1. Identify the Two Right Triangles: Consider $\Delta OQP$ and $\Delta ORP$.
2. Verify Right Angles: By Theorem 10.1 (radius is perpendicular to tangent at point of contact): $\angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ$ Therefore, both $\Delta OQP$ and $\Delta ORP$ are right-angled triangles.
3. Compare Corresponding Elements: In right triangles $\Delta OQP$ and $\Delta ORP$:
   • Hypotenuse: $OP = OP$ (Common hypotenuse to both triangles)
   • Side (Radii): $OQ = OR$ (Radii of the same circle)
   • Right Angles: $\angle OQP = \angle ORP = 90^\circ$
4. Apply RHS Congruence Criterion: $\mathbf{\Delta OQP \cong \Delta ORP \quad (\text{by RHS Congruence})}$
5. Apply CPCT (Corresponding Parts of Congruent Triangles): $PQ = PR$ Hence, proved.

3. Crucial Corollaries of Theorem 10.2 (CBSE High-Frequency)

Because $\Delta OQP \cong \Delta ORP$, three vital corollaries follow directly by CPCT:

                                      Q
                                     /|
                                  1 / |
                             O ----+  |
                                  2 \ |
                                     \|
                                      R
                                       \ 3
                                        P (Angle bisected: ∠3 = ∠4)
                                       / 4

Corollary 1: Tangents Subtend Equal Angles at the Center

$\mathbf{\angle 1 = \angle 2 \implies \angle POQ = \angle POR}$

Corollary 2: Line from Center Bisects the Angle Between Tangents

$\mathbf{\angle 3 = \angle 4 \implies \angle OPQ = \angle OPR}$ The line segment joining the center of the circle to an external point is the angle bisector of the angle between the two tangents.

Corollary 3: Tangent Angle and Center Angle Are Supplementary

In quadrilateral $OQPR$:

• The sum of all four interior angles is $360^\circ$.
• Since $\angle OQP = 90^\circ$ and $\angle ORP = 90^\circ$, their sum is $180^\circ$.
• Therefore, the remaining two angles must sum to $180^\circ$: $\mathbf{\angle QPR + \angle QOR = 180^\circ}$

<u>The angle between two tangents drawn from an external point and the angle subtended by the line segment joining the points of contact at the center are SUPPLEMENTARY ($180^\circ$).</u>

4. Solved CBSE Board Examination Problems

Solved Example 1: Supplementary Tangent Angle Calculation

Problem: Two tangents $TP$ and $TQ$ are drawn to a circle with center $O$ from an external point $T$. If $\angle POQ = 110^\circ$, find $\angle PTQ$.

Solution:

1. In quadrilateral $OPTQ$, $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$ (by Theorem 10.1).
2. By Corollary 3: $\angle PTQ + \angle POQ = 180^\circ$ $\angle PTQ + 110^\circ = 180^\circ \implies \angle PTQ = 180^\circ - 110^\circ = \mathbf{70^\circ}$
3. Therefore, <u>$\angle PTQ = 70^\circ$</u>.

Solved Example 2: Concentric Circles Chord Problem (NCERT Classic)

Problem: Two concentric circles are of radii $5\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

                                      O (Center)
                                     /|
                         Radius R=5 / | Radius r=3
                                   /  |
    A --------------------------- P --+-------------------------- B (Chord)
                                 (Point of Contact)

Solution:

1. Analyze the Geometry:
   • Let the common center of the concentric circles be $O$.
   • Let $AB$ be the chord of the larger circle (radius $R = 5\text{ cm}$) which touches the smaller circle (radius $r = 3\text{ cm}$) at point $P$.
   • Since $AB$ touches the smaller circle at $P$, $AB$ is a tangent to the smaller circle.
   • By Theorem 10.1, radius $OP \perp AB$: $OP \perp AB \implies \angle OPB = 90^\circ$
2. From Class 9 Geometry: The perpendicular from the center of a circle to a chord bisects the chord: $AP = PB \implies AB = 2 \times PB$
3. In Right Triangle $\Delta OPB$: $OB^2 = OP^2 + PB^2$ $5^2 = 3^2 + PB^2$ $25 = 9 + PB^2 \implies PB^2 = 16 \implies PB = 4\text{ cm}$
4. Calculate Total Chord Length: $AB = 2 \times PB = 2 \times 4 = \mathbf{8\text{ cm}}$
5. Therefore, <u>the length of the chord of the larger circle is $8\text{ cm}$</u>.

5. Summary and Examination Tips

Exam Tip: When proving Theorem 10.2 in board exams, explicitly state the RHS Congruence Criterion (Right angle, Hypotenuse $OP$, Side radius $OQ = OR$). Forgetting to write the congruence criterion can lose you half a mark!

Common Mistake: Assuming tangents can be drawn from inside a circle. Remember: Points inside a circle have ZERO tangents!