Question

Find the equation of circle passing through the points (1,-4) ,(5,2) and having its centre on the line x-2y+9=0

Answer 1

Let the equation of the circle be

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

since it passes through (1,-4) and (5,2), we get

$1^2+(-4)^2+2g(1)+2f(-4)+c=0$

$1+16+2g-8f+c=0$

$\bf\,2g-8f+c=-17$........(1)

and

$5^2+(2)^2+2g(5)+2f(2)+c=0$

$25+4+10g+4f+c=0$

$\bf\,10g+4f+c=-29$........(2)

(1)-(2) gives

$-8g-12f=12$

$\bf\,-2g-3f=3$..........(3)

since the centre (-g,-f) lies on the line x-2y+9=0, we get

$-g+2f+9=0$

$\implies\bf\:g=2f+9$.......(4)

using (4) in (3), we get

$-2(2f+9)-3f=3$

$-4f-18-3f=3$

$-7f=21$

$\implies\bf\:f=-3$

put f=-3 in (4), we get

$g=2(-3)+9$

$g=-6+9$

$\implies\bf\:g=3$

put f=-3 and g=3 in (1) we get

$2(3)-8(-3)+c=-17$

$6+24+c=-17$

$30+c=-17$

$\implies\bf\,c=-47$

$\therefore$The equation of the required circle is

$x^2+y^2+2(3)x+2(-3)y+(-47)=0$

$\implies\boxed{\bf\,x^2+y^2+6x-6y-47=0}$