Question

Find the equation of circle which circumscribes the triangle forms by the lines

$x+y+3=0,x-y+1=0$ and $x=3$

Answer 1

Lines $x+y+3=0$ and $x-y+1=0$ are perpendicular.

Hence, these three lines makes a right-angle triangle.

Let $A$ is point of intersection of lines

$x+y+3=0andx=3$

$\Rightarrow x+y+3=0$

$\Rightarrow 3+y+3=0$

$\Rightarrow y=-6$

$\therefore A(3,-6)$

Let $B$ is point of intersection of lines

$x-y+1=0andx=3$

$\Rightarrow x-y+1=0$

$\Rightarrow 3-y+1=$

$\Rightarrow y=4$

$\therefore B(3,4)$

Equation of circle taking $AB$ as a diameter

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\Rightarrow (x-3)(x-3)+(y+6)(y-4)=0$

$\Rightarrow x^2-6x+9+y^2+2y-24=0$

$\Rightarrow x^2+y^2-6x+2y-15=0$

Final answer:
Equation of circle is

$x^2+y^2-6x+2y-15=0$