Question

Find an equation of the circle that satisfies the given conditions.

Centre $(-1,5)$; passes through$(-4,-6)$

Short answer

Equation of the circle is ${(x+1)}^2+{(y-5)}^2=130$.

Step-by-step solution

Step 1. Determining general Equation of circle.

An equation of the circle with centre $(h,k)$and radius is${(x-h)}^2+{(y-k)}^2=r^2$

This is called the standard form for the equation of the circle.

Step 2. Given centers and points of a circle.

Center is$(-1,5)$ and passes through$(-4,-6)$

Step 3. Determining the equation of a circle.

Here, $\begin{array}{l}h=-1,k=5and\\ x=-4,y=-6\end{array}$

So,

$$\begin{gathered} {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2 \\ or{\left(-4+1\right)}^2+{\left(-6-5\right)}^2=r^2 \\ or,9+121=r^2 \\ or,r^2=130 \\ or,r=\sqrt{130} \end{gathered}$$

Now, the equation of a circle is

$\begin{array}{c}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ or{\left(x-1\right)}^2+{\left(y-5\right)}^2=130\end{array}$