Question

P is the point in the exterior of ⦿ (O,r) and the tangents from P to the circle touch the circle at X and Y.
(1) Find OP, if r = 12, XP = 5
(2) Find m∠XPO, if m∠XOY = 110
(3) Find r, if OP = 25 and PY = 24
(4) Find m∠XOP, if m∠XPO = 80

Answer 1

i ) OX = r = 12 , XP = 5 , OP = ?

Since OX is perpendicular to XP

In ∆OPX , we have

OX² = XP² + OP²

r² = XP² + OP²

12² = 5² + OP²

144 - 25 = OP²

OP² = 119

OP = √119

ii ) m<XPO = ? , m<XOY = 110°

m<XPO + m<XOY = 180°

m<XPO + 110° = 180°

m<XPO = 180° - 110°

= 70°

iii ) OP = 25 , PY = 24 , r = OY = ?

OY is perpendicular to PY

In ∆OYP ,

OP² = PY² + OY²

25² = 24² + r²

r² = 25² - 24²

r² = ( 25 + 24 )( 25 - 24 )

r² = 49

r = √49

r = 7

iv ) m<XOP = ? , m<XPO = 80°

In ∆OXP , m<PXO = 90°

m<XOP + m<XPO = 90°

m<XOP + 80° = 90°

m<XOP = 90° - 80°

m<XOP = 10°

I hope this helps you.

: )