Chapter 12, Problem 33CR

To determine

To find: The value of $x$ which is a part of a secant.

Answer to Problem 33CR

The value of x is 3.

Explanation of Solution

Given Information:

Length of secant $PR=9+x$

Length of tangent $QR=6$

Formula used:

By tangent-secant property for secant $ABC$ and tangent $AQ$

  $AC\times AB=A{Q}^2$

Calculation:

Let secant $PR$ intersects the circle at point $O$

  $\therefore PO=9\text{and}\text{}OR=x$

Now, $POR$ is a secant and $QR$ is a tangent

Therefore, by using the tangent-secant property we have

  $PR\times OR=Q{R}^2$

  $\begin{array}{l}\therefore \left(x+9\right)\times x=6^2\\ \therefore x^2+9x=36\\ \therefore x^2+9x-36=0\\ \therefore x^2+12x-3x-36=0\\ \therefore \left(x+12\right)\left(x-3\right)=0\\ \therefore x=-12\text{or}x=3\end{array}$

Now, $x$ can’t be negative as it is the length.

Thus, we have $x=3$

Hence, the value of $x$ is 3.