Question

P is the point in the exterior of $ \odot \,\left( {O,\,r} \right) $ 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\angle XPO, $ if $ m\angle XOY = 110 $
3. Find r, if $ OP = 25 $ and $ PY = 24 $
4. Find $ m\angle XOP, $ if $ m\angle XPO = 80 $

Answer 1

Hint: To find different values we use the concept that at point of contact radius is perpendicular to tangent. Therefore the triangle formed on joining external point to centre and point of contact of tangent to centre is a right angle triangle and we know that in the right triangle Pythagoras theorem holds true using this we can find different parts of the triangle.

Complete step-by-step answer:
(1)

For part (1) in the right angle triangle PXO right angle at X, both perpendicular and base are given. We use Pythagora's theorem to find the hypotenuse or third side of triangle.
OX = $ 12 $ and PX = $ 5 $
By Pythagoras theorem. We have
 $ O{P^2} = O{X^2} + P{X^2} $
Substituting values in above we have,
 $
  {\left( {OP} \right)^2} = {(12)^2} + {\left( 5 \right)^2} \\
   \Rightarrow {(OP)^2} = 144 + 25 \\
   \Rightarrow {(OP)^2} = 169 \\
   \Rightarrow OP = 13 \;
  $

(2)

Here, in problem (2) $ \angle XOY = {110^0} $ in triangle PXO and triangle PYO
OX = OY (radius of circle)
OP = OP (common side)
PX= PY (length of tangents from external point are equal)
Therefore, $ \Delta PXO \cong \Delta PYO $
 $
   \Rightarrow \angle XOP = \dfrac{1}{2}\angle XOY \\
   \Rightarrow \angle XOP = \dfrac{1}{2}(110) \\
   \Rightarrow \angle XOP = {55^0} \\
  $
Now in triangle PXO, $ \angle X = {90^0} $
Therefore by angle sum property of triangle we have
 $
  \angle XPO + \angle X + \angle XOP = {180^0} \\
   \Rightarrow \angle XPO + {90^0} + {55^0} = {180^0} \\
   \Rightarrow \angle XPO + {145^0} = {180^0} \\
   \Rightarrow \angle XPO = {180^0} - {145^0} \\
   \Rightarrow \angle XPO = {35^0} \;
  $
Therefore, from above we see that the measure of angle XPO is $ {35^0} $ .

(3)

Here, in problem (3)
Triangle OPY is a right angle at Y.
And also it is given $ OP = 25,\,PY = 24 $
Using Pythagoras theorem we have
 $ O{P^2} = P{Y^2} + O{Y^2} $
Substituting values in above we have,
 $
  {(25)^2} = {(24)^2} + {(OY)^2} \\
   \Rightarrow 625 = 576 + {r^2} \\
   \Rightarrow 625 - 576 = {r^2} \\
   \Rightarrow {r^2} = 49 \\
   \Rightarrow r = 7 \;
  $
Therefore, from above we see that the radius of a circle is $ 7 $ .

(4)

Here, in problem (4) we have
 $ m\angle XPO = {60^o} $
And triangle XPO is a right angle triangle at X.
Therefore in $ \Delta XPO $ by angle sum property. we have,
 $
  \angle X + \angle XPO + \angle XOP = {180^0} \\
   \Rightarrow {90^0} + {60^0} + \angle XOP = {180^0} \\
   \Rightarrow {150^0} + \angle XOP = {180^0} \\
   \Rightarrow \angle XOP = {180^0} - {150^0} \\
   \Rightarrow \angle XOP = {30^0} \;
  $
Therefore, from above we see that the value of $ \angle XOP\,\,is\,\,{30^o}. $

Note: We know that the length of tangents drawn from an external point to a circle are equal and also the radius of a circle at the point of contact of the tangent makes an angle of $ {90^0}. $ Hence, the triangle so formed will be a right angle triangle. So, by using Pythagora's theorem and properties of triangles we can find respective parts of the triangle.