SOLUTION: Decide whether or not this equation has a circle as its graph. If it does, give the center and the radius. 8x^2+40x+8y^2+24y-60=0

Algebra ->  Coordinate-system -> SOLUTION: Decide whether or not this equation has a circle as its graph. If it does, give the center and the radius. 8x^2+40x+8y^2+24y-60=0      Log On


   

Question 397952: Decide whether or not this equation has a circle as its graph. If it does, give the center and the radius.
8x^2+40x+8y^2+24y-60=0
Found 2 solutions by Tatiana_Stebko, lwsshak3:
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
(8x^2+40x)+(8y^2+24y)-60=0
8(x^2+5x)+8(y^2+3y)-60=0
8((x^2+2*(5/2)x+(5/2)^2)-(5/2)^2)+8((y^2+2*(3/2)y+(3/2)^2)-(3/2)^2)-60=0(/8)
(x+(5/2))^2-(25/4)+(y+(3/2))^2-(9/4)-7.5=0
(x+2.5)^2+(y+1.5))^2=16
%28x-a%29%5E2%2B%28y-b%29%5E2=R%5E2 has a circle as a graph with center in (a;b) and radius R
so (x+2.5)^2+(y+1.5)^2=16 has a circle as a graph with center in (-2.5;-1.5) and radius R=4

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
8x^2+40x+8y^2+24y-60=0
standard form for a circle
(x-h)^2+(y-k)^2=r^2, with (h,k) being the coordinates of the center, and r=radius.
completing squares
8(x^2+5x+25/4)+8(y^2+3y+9/4)=60+50+18=128
divide by 8
(x+5/2)^2+(y+3/2)^2=16
center(-5/2,-3/2)
r^2=16
r=4
ans: this is a circle with center at (-5/2,-3/2), and radius=4