Question

How do you find the radius and center of a circle with the equation `5x^2+5y^2+60x+40y+215=0`?

Answer 1

Center `(-6, -4)` and Radius `=3`

Explanation:

From the give equation `5x^2+5y^2+60x+40y+215=0`,
we can see numbers that are divisible by 5.
We divide first by 5 to simplify the solution and the equation becomes

`x^2+y^2+12x+8y+43=0`,
We now perform "completing the square method"

`x^2+y^2+12x+8y+43=0`,
`x^2+12x+y^2+8y=-43" " "`arranging the terms first

then add the required constants 36 and 16 on both sides of the equation
`x^2+12x+36+y^2+8y+16=-43+36+16" " "`
so that "Perfect Square Trinomials" will appear
`(x^2+12x+36)+(y^2+8y+16)=-43+36+16" " "`
and then
`(x+6)^2+(y+4)^2=9`
Convert now to "Center Radius Form"
`(x--6)^2+(y--4)^2=3^2`
We now have the Center (h, k) and Radius r by inspection

Center `(-6, -4)` and Radius `=3`

graph of `5x^2+5y^2+60x+40y+215=0`
graph{5x^2+5y^2+60x+40y+215=0[-20,5,-10,2]}

God bless...I hope the explanation is useful...