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

Center $\left(- 6 , - 4\right)$ and Radius $= 3$

#### Explanation:

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

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

${x}^{2} + {y}^{2} + 12 x + 8 y + 43 = 0$,
${x}^{2} + 12 x + {y}^{2} + 8 y = - 43 \text{ " }$arranging the terms first

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

Center $\left(- 6 , - 4\right)$ and Radius $= 3$

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

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