Chapter C, Problem 1E

To determine

To find: The equation of the circle.

Answer to Problem 1E

The equation of the circle is $x^2+y^2-6x+2y=15$.

Explanation of Solution

Given:

The center of the circle is $\left(-2,-8\right)$.

The radius of the circle is 10.

Property used:

The equation of the circle passes through the point $\left(x_1,y_1\right)$ and the radius r is, ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=r^2$.

Calculation:

Consider the point $\left(x_1,y_1\right)=\left(3,-1\right)$ and the radius $r=5$.

Substitute the values in the general equation of the circle.

$\begin{array}{l}{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=r^2\\ {\left(x-3\right)}^2+{\left(y-\left(-1\right)\right)}^2=5^2\\ {\left(x-3\right)}^2+{\left(y+1\right)}^2=25\\ x^2+9-6x+y^2+1+2y=25\end{array}$

Simplify further as,

$\begin{array}{l}{x}^2+y^2-6x+2y+10=25\\ x^2+y^2-6x+2y=25-10\\ x^2+y^2-6x+2y=15\end{array}$

Thus, the equation of the circle is $x^2+y^2-6x+2y=15$.