Question

Find the equation of the circle which passes through the points $(2,-2)$, and $(3,4)$ and whose centre lies on the line $x+y=2$.

Answer 1

Let the equation of the circle be

$x^2+y^2+2gx+2fy+c=0$

As it passes through the point $(2,-2)$ and $(3,4)$

$4+4+4g-4f+c=0$ and $9+16+6g+8f+c=0$

$8+4g-4f+c=0$ and $25+6g+8f+c=0$

Substracting the two equations, we get

$17+2g+12f=0$

As the centre is $(-f,-g)$

$-f-g=2$

$f+g=-2$

$17-4+10f=0$

$f=-1.3$

$g=-0.7$

Equation of the circle is

$x^2+y^2-1.4x-2.6y+c=0$

$4+4+5.2-2.8+c=0$

$c=-10.4$

$x^2+y^2-1.4x-2.6y-10.4=0$