Question

The equation of the circumcircle of the triangle formed by the lines $xy-3x-2y+6=0$ and $x+y=0$ is

• A. $(2x-1{)}^2+(2y-1{)}^2=50$
• B. $(2x+1{)}^2+(2y-1{)}^2=50$
• C. $(2x+1{)}^2+(2y+1{)}^2=50$
• D. $(x-1{)}^2+(2y+1{)}^2=50$

Answer 1

The correct option is B $(2x+1{)}^2+(2y-1{)}^2=50$

Given pair of lines
$xy-3x-2y+6=0$
$\Rightarrow (x-2)(y-3)=0$
$\Rightarrow x-2=0\cdots \left(1\right)$
$y-3=0\cdots \left(2\right)$
Here both lines are perpendicular.
Also, $x+y=0\cdots \left(3\right)$
Solving $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$P(-3,3),Q(2,-2),A(2,3)$


$△PAQ$ is right angled at $A$
Then $PQ$ will be the diameter.
Centre will be the mid point of $PQ.$
Centre $=(\frac{-3+2}{2},\frac{3-2}{2})=(-\frac{1}{2},\frac{1}{2})$
and $PQ=\sqrt{(2+3{)}^2+(-2-3{)}^2}=\sqrt{50}=5\sqrt{2}$
Radius $=\frac{PQ}{2}=\frac{5}{\sqrt{2}}$
$\therefore$ Equation of the circumcircle is
${(x+\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=\frac{25}{2}$
$\Rightarrow (2x+1{)}^2+(2y-1{)}^2=50$