How do you find the radius of the circle x^2 + y^2 - 4x + 6y - 12 = 0x2+y2−4x+6y−12=0?

1 Answer
Cesareo R.
May 26, 2016

r=5r=5 in (x-2)^2+(y+3)^2=5^2(x−2)2+(y+3)2=52

Explanation:

The circle equation can be arranged as

(x-x_0)^2+(y-y_0)^2=r^2(x−x0)2+(y−y0)2=r2

in which x_0,y_0x0,y0 are the center coordinates and rr the radius.
Expanding and comparing

x^2+y^ 2-2x_0 x-2 y_0 y +x_0^2+y_0^2-r^2 = 0x2+y2−2x0x−2y0y+x20+y20−r2=0
x^2+y^2-4x+6y-12=0x2+y2−4x+6y−12=0

so we have

{(-2x_0=-4),(-2y_0=6),(x_0^2+y_0^2-r^2=-12):}

Solving for x_0,y_0,r easily we obtain

x_0=2,y_0=-3,r=5

so the equation reads

(x-2)^2+(y+3)^2=5^2

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