Chapter 11.1, Problem 48SPR

To determine

To find: The equation of the circle.

Answer to Problem 48SPR

The required equation of the circle is ${\left(x+3\right)}^2+{\left(y+10\right)}^2=144$ .

Explanation of Solution

Given:

The centre of the circle is at $\left(-3,-10\right)$ and the diameter of the circle is $d=24$ .

Calculation:

Consider the radius of the circle is evaluated as,

  $\begin{array}{c}r=\frac{d}{2}\\ =\frac{24}{2}\\ =12\end{array}$

Consider the formula for the equation of the circle is of the form as,

  ${\left(x-h\right)}^2+\left(y-k^2\right)=r^2$

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

The centre of the circle is at,

  $\left(h,k\right)=\left(-3,-10\right)$

From, the general equation of the circle, the equation of the circle is,

  $\begin{array}{l}{\left(x-h\right)}^2+\left(y-k^2\right)=r^2\\ {\left(x-\left(-3\right)\right)}^2+\left(y-{\left(-10\right)}^2\right)={\left(12\right)}^2\\ {\left(x+3\right)}^2+{\left(y+10\right)}^2=144\end{array}$