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

Solution

Given: AP and AQ are two tangents from a point A to a circle c(0,r)

To prove: AP = AQ

construction: Join OP, OQ and OA.

Proof:

$In ∆ OPA and ∆ OQA, ∠ OPA = ∠ OQA = 90 ° . . . . . . . (Tangent at any point of a circle is perpendicular to the radius through the point of contact) OP = OQ . . . . . . . . . (Radii of a circle) OA = OA . . . . . . . . . (Common)$

Hence, by RHS- criterion of congruence, e have

$∆ OPA ≅ ∆ OQA \Rightarrow AP = AQ . . . . . . . . (c . p . c . t .)$

For Daily Free Study Material Join wiredfaculty

Circles

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

Some More Questions From Circles Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction