Chapter 13.3, Problem 8PSB

To determine

To show: The point(s) of intersection of the given circle and hyperbola.

Answer to Problem 8PSB

The point(s) of intersection $\left(1\right)\&\left(2\right)$ are $\left(4,3\right),\left(-4,3\right),\left(4,-3\right),\left(-4,-3\right)$ .

Explanation of Solution

Given:

The given equation of circle is:

  $x^2+y^2=25\left(1\right)$

The given equation of hyperbola is:

  $x^2+y^2=7\left(2\right)$

Calculation:

The point(s) of intersection of equation $\left(1\right)\&\left(2\right)$ is calculated as:

  $\begin{array}{c}2x^2=32\left[\text{add equation}\left(1\right)\text{\&equation}\left(2\right)\right]\\ x=\pm 4\\ y^2=25-16\left[\text{put }x=\pm 4\text{ in equation }\left(1\right)\right]\\ y=\pm 3\end{array}$

Therefore the point(s) of intersection $\left(1\right)\&\left(2\right)$ are $\left(4,3\right),\left(-4,3\right),\left(4,-3\right),\left(-4,-3\right)$ .