How do you write the equation for a circle that passes through the points at (4,5), (-2,3) and (-4,-3)?

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Answer 1 · 2016-06-26T12:06:23

$x^2+y^2-6x+4y-37=0$

let's substitute each coordinate of all points in the general equation of the circle:

$x^2+y^2+ax+by+c=0$

so that you have three equations with unknown variables a,b,c:

$4^2+5^2+4a+5b+c=0$ and
$(-2)^2+3^2-2a+3b+c$ and
$(-4)^2+(-3)^2-4a-3b+c=0$

let's solve in a few steps:
$4a+5b+c+41=0 and -2a+3b+c+13=0 and -4a-3b+c+25=0$

$c=-4a-5b-41 and$ (by substituting c) $-2a+3b-4a-5b-41+13=0 and -4a-3b-4a-5b-41+25=0$

$-6a-2b-28=0 and -8a-8b-16=0$

that is equivalent to:

$3a+b+14=0 and a+b+2=0$

By subtracting you obtain:

$2a+12=0$

$a=-6$

so, by substituting the obtained value of a:

$-6+b+2=0$

$b=4$

so

$c=-4(-6)-5(4)-41=24-20-41=-37$

The equation of the circle is:

$x^2+y^2-6x+4y-37=0$