Question

Find the equation of the circle passing through the points
$(1,2),(3,-4)$ and $(5,-6)$

Answer 1

$x^2+y^2+2gy+2fy+c=0$

passing from $(1,2)$ then

$1+4+2g+4f+c=0$

$=2g+4f+c=-5$.....................$\left(1\right)$

from$(3,-4)$

$9+16+8f+c=0$

$6g-8f+c=-25$...............$\left(2\right)$

from$(5,-6)$

$25+36+10g-12f+c=0$

$10g-12f+c==61$............$\left(3\right)$

on substituting equation $\left(1\right)$ and $\left(2\right)$

hence, we have

$-4g+12f=20$

$-g+3f=5$..................$\left(4\right)$

on substituting equation $2$ and $3$

we get

$-4g+4f=36$

$-g+f=9$............$\left(5\right)$

on substituting equation $4$ and $5$

we get

$2f=-4$

$f=-2$

$\therefore g==11$

and $c=25$

putting in the equation

we get

$x^2+y^2-22x-4f+25=0$