Chapter 10.4, Problem 11PSB

a)

To determine

Radius of the circle.

Answer to Problem 11PSB

  $\text{Radius = 2}\sqrt{\text{13}}\text{ units}$

Explanation of Solution

Given: $\text{Centre at point P(}-2,5\text{) and point Q(4, 9)}\text{.}$

  

Calculation:

  $\text{Radisu = QP = }\sqrt{{(4+2)}^2+{(9-5)}^2}=\sqrt{36+16}=2\sqrt{13}\text{units}$

b)

To determine

Slope of tangent at Q.

Answer to Problem 11PSB

  $\text{Slope of tangent = }\frac{-3}{2}$

Explanation of Solution

Calculation:

  $\begin{array}{l}\text{As radius is perpendicular to tangent}\text{.}\\ {\text{Let slope of PQ is m}}_{\text{1}}{\text{ and slope of tangent at Q is m}}_{\text{2}}.\\ \therefore \text{PQ is perpendicular to tangent at Q}\text{.}\\ \Rightarrow m_1\times m_2=-1\\ m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{9-5}{4+2}=\frac{4}{6}=\frac{2}{3}\\ m_2=\frac{-3}{2}\end{array}$