Q. 15.0( 1 Vote )

# P is the point in

Given that P is the point in exterior of circle (O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Here OX is the radius of the circle and PX is the tangent. OX PX

Given r = 12 = OX and XP = 5

For right angled ΔPXO,

OP2 = PX2 + OX2

= 52 + 122

= 25 + 144

= 169

OP2 = 169

OP = 13

(2) Given XOY = 110° Here XPY and XOY are supplementary angles.

We know that two angles are supplementary when they add upto 180°.

XPY + XOY = 180°

XPY + 110° = 180°

XPY = 70°

Also, OP is a bisector pf XPY

XPO = 1/2 XPY = 1/2 (70°) = 35°

(3) OY is the radius of the circle and PY is a tangent.

OY PY

Now, OY = r,

Given that OP = 25 and PY = 24 In right angled ΔPYO,

By Pythagoras Theorem,

OP2 = OY2 + PY2

252 = r2 + 242

r2 = 625 – 576

r2 = 49

r = 7

(4) Given XPO = 80°

Here, OX is the radius of the circle and PX is a tangent. OX PX

Also, XOP and XPO are complementary angles.

We know that two angles are complementary when they add up to 90°.

XOP + XPO = 90°

XOP + 80° = 90°

XOP = 10°

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