Question

Find the centre of circle passing through $A(0,1),B(2,3),C(-2,5)$ is

Answer 1

Let the circle be $x^2+y^2+Dx+Ey+F=0$ .....$\left(1\right)$

$A(0,1)$ passes through equation$\left(1\right)$

$\Rightarrow 0+1+0+E+F=0$

$\Rightarrow E+F=-1$ .....$\left(1\right)$

$B(2,3)$ passes through equation$\left(1\right)$

$\Rightarrow 4+9+2D+3E+F=0$

$\Rightarrow 2D+3E+F=-13$ ....$\left(3\right)$

$C(-2,5)$ passes through equation$\left(1\right)$

$\Rightarrow 4+25-2D+5E+F=0$

$\Rightarrow -2D+5E+F=-29$ ....$\left(4\right)$

$\left(3\right)\Rightarrow 2D+3E+F=-13$

$\Rightarrow 2D+2E+E+F=-13$

$\Rightarrow 2D+2E-1=-13$ from $\left(2\right)$

$\Rightarrow 2D+2E=-13+1=-12$

$\Rightarrow D+E=-6$ ........$\left(5\right)$

$\left(4\right)\Rightarrow -2D+4E+E+F=-29$

$\Rightarrow -2D+4E-1=-29$ from $\left(2\right)$

$\Rightarrow -2D+4E=-29+1=-28$

$\Rightarrow -D+2E=-14$ .....$\left(6\right)$

Solving equations $\left(5\right)$ and $\left(6\right)$ we get

$\left(5\right)+\left(6\right)\Rightarrow D+E-D+2E=-6+(-14)=-20$

$\Rightarrow 3E=-20$

$\therefore E=\frac{-20}{3}$

Substituting for $E=\frac{-20}{3}$ in eqn$\left(5\right)$ we get

$D=-6-E=-6-\left(\frac{-20}{3}\right)=-6+\frac{20}{3}=\frac{-18+20}{3}=\frac{2}{3}$

Substituting the values of $D=\frac{2}{3}$ and $E=\frac{-20}{3}$ in

eqn$\left(4\right)$ we get

$\left(4\right)\Rightarrow -2D+5E+F=-29$

$\Rightarrow -2\left(\frac{2}{3}\right)+5\left(\frac{-20}{3}\right)+F=-29$

$\Rightarrow \left(\frac{-4}{3}\right)+\left(\frac{-100}{3}\right)+F=-29$

$\Rightarrow F=-29-\left(\frac{-4}{3}\right)-\left(\frac{-100}{3}\right)$

$\Rightarrow F=-29+\frac{4}{3}+\frac{100}{3}=-29+\frac{104}{3}=\frac{-87+104}{3}=\frac{17}{3}$

$\therefore D=\frac{2}{3},E=\frac{-20}{3},F=\frac{17}{3}$

$\therefore$ Equation of the circle is $x^2+y^2+Dx+Ey+F=0$ becomes $x^2+y^2+\frac{2}{3}x+\frac{-20}{3}y+\frac{17}{3}=0$

is of the form $x^2+y^2+2gx+2fy+c=0$

Centre is $(-g,-f)$

$\Rightarrow 2g=\frac{2}{3}\Rightarrow g=\frac{1}{3}$

$\Rightarrow 2f=\frac{-20}{3}\Rightarrow f=\frac{-10}{3}$

Hence centre is at $(\frac{-1}{3},\frac{10}{3})$