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

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Answer 1 · 2016-05-26T14:19:48

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

The circle equation can be arranged as

$(x-x_0)^2+(y-y_0)^2=r^2$

in which $x_0,y_0$ are the center coordinates and $r$ the radius.
Expanding and comparing

$x^2+y^ 2-2x_0 x-2 y_0 y +x_0^2+y_0^2-r^2 = 0$
$x^2+y^2-4x+6y-12=0$

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

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

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