Question

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

Answer 1

Let the equation of circle be-

$x^2+y^2+2gx+2fy+c=0.....\left(1\right)$

Given that the circle passes through the points $(3,4),(3,2)$ and $(1,4)$.

Therefore,

${\left(3\right)}^2+{\left(4\right)}^2+2g\left(3\right)+2f\left(4\right)+c=0$

$\Rightarrow 9+16+6g+8f+c=0$

$\Rightarrow 6g+8f+c+25=0.....\left(2\right)$

${\left(3\right)}^2+{\left(2\right)}^2+2g\left(3\right)+2f\left(2\right)+c=0$

$\Rightarrow 9+4+6g+4f+c=0$

$\Rightarrow 6g+2f+c+13=0.....\left(3\right)$

${\left(1\right)}^2+{\left(4\right)}^2+2g\left(1\right)+2f\left(4\right)+c=0$

$\Rightarrow 1+16+2g+8f+c=0$

$\Rightarrow 2g+8f+c+17=0.....\left(4\right)$

Subtracting equation $\left(3\right)$ from $\left(2\right)$, we have

$(6g+8f+c+25)-(6g+4f+c+13)=0$

$6g+8f+c+25-6g-4f-c-13=0$

$\Rightarrow f=\frac{-12}{4}=-3$

Now subtracting equation $\left(4\right)$ from $\left(2\right)$, we have

$(6g+8f+c+25)-(2g+8f+c+17)=0$

$6g+8f+c+25-2g-8f-c-17=0$

$\Rightarrow g=\frac{-8}{4}=-2$

Now substituting the value of $g$ and $f$ in equation $\left(2\right)$, we have

$6(-2)+8(-3)+c+25=0$

$-12-24+c+25=0$

$\Rightarrow c=11$

Substituting the values of $g,f$ and $c$ in equation $\left(1\right)$,

we get

$x^2+y^2+2(-2)x+2(-3)y+\left(11\right)=0$

$x^2+y^2-4x-6y+11=0$

Hence the equation of circle passing through the given points is $x^2+y^2-4x-6y+11$.