Question:
Find the equation of the circle which passes through the points (1, 3) and (2, - 1), and has its centre on the line 2x + y – 4 = 0.
Solution:
The equation of a circle: $x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i)
Putting $(1,3) \&(2,-1)$ in (i)
$2 g+6 f+c=-10 . .($ ii $)$
$4 g-2 f+c=-5 . .($ iii $)$
Since the centre lies on the given straight line, $(-g,-f)$ must satisfy the equation as
$-2 g-f-4=0 \ldots$ (iv)
Solving, $f=-1, g=-1.5, c=-1$
The equation is $x^{2}+y^{2}-3 x-2 y-1=0$