Question

Find the equation of circle which passes through the points $(5,0)$ and $(1,4)$ and whose centre lie on the $x+y-3=0$.

Answer 1

Let the equation of circle is $x^2+y^2+2gx+2fy+c=0$

Centre $=(-g,-f)$

Given centre lies on $x+y-3=0$

i.e; $-g-f-3=0$

$g+f+3=0---\left(1\right)$

Given circle is passing through $(5,0)$ and $(1,4)$ so

$25+0+10x+0+c=0---\left(2\right)$

$1+16+2g+8f+c=0---\left(2\right)$

eg $\left(2\right)-\left(3\right)$ we get

$8+8g=8f\to 1+g=f$

Sub ${}^{\prime }{f}^{\prime }$ in $\left(1\right)$ $g+1+g+3=0$

$g+f+3=0$ $g=-2$

$f=-1$ $c=-5$

$\therefore$ Equation of circle is

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