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

1 Answer

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...