Question

Find the equation of the circle.What is the center and radius of the circle?Be sure to show all work and explain your answer.

Answer 1

`(x+4)^(2)+(y-1)^(2)=30`

Center (4, -1), r = √30

Explanation:

We have an equation in its general form, we must convert it to its standard form of equation of the circle to determine the center and radius, like this:

`x^2+y^2+8x-2y-13=0`

In the following way, we solve correctly

`\begin{gathered} \text{ The equation of the circle:} \\ \\ \left(x−h\right)^2+\left(y−k\right)^2=r^2\: \\ \\ (h,k)\text{ is the center and r is the radius } \\ \\ \text{ Therefore:} \\ \\ x^2+y^2+8x-2y-13=0 \\ \\ x^2+y^2+8x-2y=13 \\ \\ \left(x^2+8x\right)+\left(y^2-2y\right)=13 \\ \\ \left(x^2+8x+16\right)+\left(y^2-2y+1\right)=13+16+1 \\ \\ \left(x+4\right)^2+\left(y-1\right)^2=30 \\ \\ \text{ Thus} \\ \\ \text{ The center is \lparen-4,1\rparen} \\ \\ \text{ The radius is }√(30) \end{gathered}`

Center (4, -1), r = √30