Question

Find the equation of the circle passing through the point $(5,7),(8,1),(1,3)$.

Answer 1

Let the circle be $x^2+y^2+Dx+Ey+F=0$

$(5,7):5^2+1^2+5D+7E+F=0$

$\Rightarrow 5D+7E+F=-26$ …………….$\left(1\right)$

$(8,1):8^2+1^2+8D+E+F=0$

$\Rightarrow 8D+E+F=-65$ …………$\left(2\right)$

$(1,3):1^2+3^2+D+3E+F=0$

$\Rightarrow D+3E+F=-10$ …………$\left(3\right)$

Equation $\left(1\right)$ - equation $\left(2\right)$

$\Rightarrow 5D+7E+R-8D-E-F=-26+65$

$\Rightarrow -3D+6E=39$

$\Rightarrow -D+2E=13$ ………$\left(4\right)$

Equation $\left(2\right)$ - equation $\left(3\right)$

$\Rightarrow 8D+E+F-D-3E-F=-65+10$

$\Rightarrow 7D-2E=-55$ ………..$\left(5\right)$

Equation $\left(4\right)$ $+$ equation $\left(5\right)$

$\Rightarrow -D+2E+7D-2E=13-55$

$6D=-42$

$\therefore D=-7$

Put $D=-7$ in $\left(4\right)$

$\Rightarrow 2E=13+D=13-7=6$

$\therefore E=6/2=3$

Put $D=-7$, $E=3$ in $\left(3\right)$ we get

$7+3\times 3+F=-10$

$\Rightarrow -7+9+F=-10$

$\Rightarrow F=-10-2=-12$

$\therefore E=3,F=-12,D=-7$

The equation of the circle is $x^2+y^2-7x+3y-12=0$.