Geo4.18

Find the circle which passes through (-1,2),(3,-2) and has its centre on the line $x-2y=0\,$

Let the equation of the circle be

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

The point (-1,2) lies on the circle,hence

$1+4-2g+4f+c=0\,$

$-2g+4f+c=-5\,$

Point (3,-2) lies on the circle

$9+4+6g-4f+c=0\,$

$6g-4f+c=-13\,$

Also the centre lies on $x-2y=0\,$

Hence

$-g+2f=0\,$

$g=2f\,$

Substituting this in the first equation we get

$-2(2f)+4f+c=-5\,$

$c=-5\,$

Now

$6(2f)-4f-5=-13\,$

$8f=-8\,$

Hence f=-1, Now g=-2

Equation of the circle is

$x^2+y^2-4x-2y-5=0\,$