Question

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Hint:

In circle geometry proofs, always connect the center to tangent points — it reveals right triangles and makes congruence proofs easier.

Answer 1

Concept: We use properties of tangents and triangles:

• A tangent is perpendicular to the radius at the point of contact.
• If two right triangles have equal hypotenuse and one equal side, they are congruent (RHS congruence).

Step 1: Construct the figure.
Let a circle have center \( O \), and let \( P \) be an external point. Draw tangents \( PA \) and \( PB \) touching the circle at points \( A \) and \( B \).
Step 2: Join radii to points of contact.
Join \( OA \) and \( OB \). Since radius is perpendicular to tangent at the point of contact: \[ OA \perp PA, \quad OB \perp PB. \] Thus, \( \triangle OAP \) and \( \triangle OBP \) are right triangles.
Step 3: Compare the triangles.
In triangles \( \triangle OAP \) and \( \triangle OBP \):

• \( OA = OB \) (radii of the same circle)
• \( OP = OP \) (common side)
• \( \angle OAP = \angle OBP = 90^\circ \)

Step 4: Apply RHS congruence.
By RHS congruence rule: \[ \triangle OAP \cong \triangle OBP. \]
Step 5: Corresponding sides are equal.
Therefore, corresponding tangent lengths are equal: \[ PA = PB. \] Conclusion:
The lengths of tangents drawn from an external point to a circle are equal.