Question

In the figure, $AB$ is a diameter and $AP$ is a tangent to the circle. If $PB = 9cm$ and $PQ = 4cm$, find the length of $AP.$ What is the radius of the circle?

Answer 1

Hint: In this question use the concept of tangent secant i.e $P{A^2} = PQ \times PB$ also Pythagoras Theorem i.e. ${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}$ will be used to find the diameter of the circle.

Complete step-by-step answer:

According to the question, it is given

$

  PB = 9cm \\

  PQ = 4cm \\

$

Hence, From the figure $P{A^2} = PQ \times PB$

$

   \Rightarrow P{A^2} = 4 \times 9 = 36 \\

   \Rightarrow PA = \sqrt {36} = 6cm \\

 $

In right triangle \[APB\]$A = \angle {90^ \circ }$, So by Pythagoras Theorem

${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}$

$ \Rightarrow P{B^2} = A{P^2} + A{B^2}$

$

   \Rightarrow {6^2} + A{B^2} = {9^2} \\

   \Rightarrow A{B^2} = {9^2} - {6^2} = 81 - 36 = 45 \\

   \Rightarrow AB = \sqrt {45} = \sqrt {9 \times 5} = 3\sqrt 5 \\

$

Thus, radius of the circle $ = \dfrac{{AB}}{2} = \dfrac{{3\sqrt 5 }}{2}cm$.

Note: In such types of questions the concept of tangent secant is used i.e. if a tangent segment and a secant segment are drawn to a circle from an exterior point, the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment and here from the diagram we can see that the square of measure of tangent is $PA$ and measure of secant segment and external secant is $PQ$ and $PB$ respectively.