Question

Find the equation of the circle which is circumscribed about the triangle, whose vertices are $(-2,3),(5,2)$ and $(6,-1)$.

Answer 1

Given,the vertices of triangle $(-2,3),(5,2),(6,-1)$

General equation of the circle is given by, $x^2+y^2+Dx+Ey+F=0$

$(-2,3)\Rightarrow {(-2)}^2+3^2-2D+3E+F=0\Rightarrow -2D+3E+F=-13$.....$\left(1\right)$

$(5,2)\Rightarrow {\left(5\right)}^2+2^2+5D+2E+F=0\Rightarrow 5D+2E+F=-29$.....$\left(2\right)$

$(6,-1)\Rightarrow 6^2+{(-1)}^2+6D-1E+F=0\Rightarrow 6D-E+F=-37$.....$\left(3\right)$

Solving equations $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

Equation $\left(1\right)-\left(2\right)=-2D+3E+F-5D-2E-F=-13+29=16$

$\Rightarrow -7D+E=16$ ...........$\left(4\right)$

Equation $\left(2\right)-\left(3\right)$,we get

$5D+2E+F-6D+E-F=-29+37=8$

$\Rightarrow -D+3E=8$ .........$\left(5\right)$

Solving $\left(4\right)$ and $\left(5\right)$ we get

equation $\left(4\right)-7\times$equation$\left(5\right)$ we get

$-7D+E+7D-21E=16-56=-40$

$\Rightarrow -20E=-40$ or $E=2$

Substituting the value of $E=2$ in $\left(5\right)$ we get

$\Rightarrow -D+6=8$ or $D=-2$

Substituting $D=-2$ and $E=2$ in $\left(1\right)$ we get

$4+6+F=-13$ or $F=-23$

$\therefore D=-2,E=2,F=-23$

Required equation of circle is,

$x^2+y^2-2x+2y-23=0$