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,

⇒ OP^{2} = PX^{2} + OX^{2}

= 5^{2} + 12^{2}

= 25 + 144

= 169

⇒ OP^{2} = 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,

⇒ OP^{2} = OY^{2} + PY^{2}

⇒ 25^{2} = r^{2} + 24^{2}

⇒ r^{2} = 625 – 576

⇒ r^{2} = 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°

Rate this question :

ΔABC is an isosceGujarat Board Mathematics

P is in the exterGujarat Board Mathematics

P is the point inGujarat Board Mathematics

<img width="33" hGujarat Board Mathematics

In figure 11.24, Gujarat Board Mathematics

<img width="33" hGujarat Board Mathematics

Tangents from P, Gujarat Board Mathematics

The points of conGujarat Board Mathematics

A circle touches Gujarat Board Mathematics

<img width="33" hGujarat Board Mathematics