Q. 15.0( 1 Vote )

P is the point in

Answer :

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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