Chapter 10.8, Problem 36HP

To determine

To prove: The given angle is a right angle.

Explanation of Solution

Given information: The information is given in the following figure.

  

Formula used: $m_{xy}=\frac{y_2-y_1}{x_2-x_1}$

Proof:

Finding the slope of $\overline{AC}$ by applying the slope formula,

  $\begin{array}{l}{m}_{AC}=\frac{y-r}{x-0}\\ =\frac{y-r}{x}\end{array}$

Similarly the slope of $\overline{BC}$ ,

  $\begin{array}{l}{m}_{BC}=\frac{y+r}{x-0}\\ =\frac{y+r}{x}\end{array}$

Now, multiplication of the slopes, we get,

  $\begin{array}{l}{m}_{AC}\times m_{BC}=\frac{y-r}{x}\times \frac{y+r}{x}\\ m_{AC}\times m_{BC}=\frac{y^2-r^2}{x^2}\end{array}$

Since, the point C in on the circle,

  $\begin{array}{l}{x}^2+y^2=r^2\\ y^2-r^2=-x^2\end{array}$

Replacing $y^2-r^2$ by $-x^2$ ,

  $\begin{array}{l}\therefore m_{AC}\times m_{BC}=\frac{-x^2}{x^2}\\ \therefore m_{AC}\times m_{BC}=-1\end{array}$

Hence, both lines are perpendicular because their product of slopes is $-1$ .

  $\therefore AC\perp BC$

Therefore, $\angle ACB$ is right angle.