Chapter 9.2, Problem 36WE

To determine

To find: The equation of the circle.

Answer to Problem 36WE

The required equation is ${\left(x+2\right)}^2+y^2=16$ .

Explanation of Solution

Given information: The given points are $Center\left(-2,0\right)$ and $\left(2,0\right)$ .

Formula used: Distance formula $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ and ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: First of all, finding the length of the radius by using the distance formula.

Hence,

  $\begin{array}{l}r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{16}\\ =4\end{array}$

As we know that the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ has the center $\left(h,k\right)$ and radius $\left(r\right)$ .

  $\begin{array}{l}\therefore {\left(x+2\right)}^2+{\left(y-0\right)}^2=16\left\{\because \left(h,k\right)=\left(-2,0\right);r=4\right\}\\ \therefore {\left(x+2\right)}^2+y^2=16\end{array}$

Therefore, the required equation is ${\left(x+2\right)}^2+y^2=16$ .

The graph is given below,