Page 2 - Lesson Notes-Tangents Ch-10 (Circles)

external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

Proof-

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of
contact.

OA PA and OB PB ... (1)

In OPA and OPB:

OAP = OBP (Using (1))

OA = OB (Radii of the same circle)

OP = OP (Common side)

Therefore, ∆ OPA ∆OPB (RHS congruency criterion)

PA = PB

(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are
equal.

Example 1-
If PT is a tangent at T to a circle whose center is 0 and OP = 17 cm, OT = 8 cm. Find the length
of tangent segment PT.

Solution:
OT = radius = 8 cm

OP = 17 cm
PT = length of tangent =?
T is point of contact. We know that at point of contact tangent and radius are perpendicular.
2
2
2
∴ OTP is right angled triangle ∠OTP = 90°, from Pythagoras theorem OT + PT = OP
2
2
8 + PT = 17 2
2
2
2
PT = (17 -8 )

2