Chapter 9.2, Problem 14OE

To determine

To calculate: Find the equation of the circle.

Answer to Problem 14OE

The required equation of the circle is $x^2+y^2+6x-8y+17=0$ .

Explanation of Solution

Given information: The circle has coordinates of the center of $\left(-3,4\right)$ and the radius of $\sqrt{8}$ .

Formula used: The distance formula: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation:

As we know that the center of a circle $=\left(h,k\right)$

Now, substitute the value of the coordinates of the center and the radius of the circle into ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Hence,

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-\left(-3\right)\right)}^2+{\left(y-4\right)}^2={\left(\sqrt{8}\right)}^2\left[\because \left(h,k\right)=\left(-3,4\right)\right]\\ {\left(x+3\right)}^2+{\left(y-4\right)}^2=8\\ x^2+9+6x+y^2+16-8y=8\\ x^2+y^2+6x-8y+25-8=0\\ x^2+y^2+6x-8y+17=0\end{array}$

Therefore, the required equation of the circle is $x^2+y^2+6x-8y+17=0$ .