Question

Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on the line x - 4y = 1.

Answer 1

The circle passes through P and Q and the centre lies on
$x-4y=1\dots \left(i\right)$
The general equation of circle is
$x^2+y^2+2gx+2fy+c=0\dots \left(ii\right)$
\because P and Q lies on (ii),
so, $9+49+6g+14f+c=0\dots \left(iii\right)$
$25+25+lOg+10f+c=0\dots \left(iv\right)$
Also, centre (-g, -f) lies on (i)
$\therefore -g+4f=1\dots \left(v\right)$
Solving (iii) (iv) and (v), we get
g = 3, f = 1 and c = -90
from (ii)
The equation of circle is
$x^2+y^2+6x+2y-90=0$