Question

Find the equation of circle which circumscribes the triangle forms by the lines $2x+y-3=0,x+y-1=0$ and $3x+2y-5=0.$

Answer 1

Let $A$ is the Point of intersection of line

$2x+y-3=0andx+y-1=0$

Subtract the equation

$\Rightarrow x-2=0\Rightarrow x=2$

$\because x=2=0\Rightarrow x=2$

Let $B$ is the Point of intersection of line

$x+y-1=0and3x+2y-5=0$

Again subtracting

$\Rightarrow x-3-0\Rightarrow x=3$

$\because x=3\Rightarrow y=-2$

$\therefore B(3,-2)$

Let $C$ is the Point of intersection of line

$2x+y-3=0and3x+2y-5=0$

$\Rightarrow -x+1=0\Rightarrow x=1$

$\because x=1\Rightarrow y=1$

$\therefore C(1,1)$

Let equation of such circle

$x^2+y^2+2gx+2fy+c=0\dots \left(1\right)$

Since circle passes through the vertices of triangle ABC

For point $A(2,-1)$

$2^2+1^2+2g\times 2+2f\times (-1)+c=0$

$\Rightarrow 4g-2f+c+5=0\dots \left(2\right)$

For point B $(3,-2)$

$3^2+(-2{)}^2+2g\times 3+2f\times (-2)+c=0$

$\Rightarrow 6g-4f+c+13=0\dots \left(3\right)$

And for point $C(1,1)$

$1^2+1^2+2g\times 1+2f\times 1+c=0$

$\Rightarrow 2g+2f+c+2=0\dots \left(4\right)$

For value of $g,f$ and $c$ we need to solve above three equations.

Subtract equation (2) and (3)

$\Rightarrow -2g+2f-8=0\dots \left(5\right)$

Again, subtracting the equation (2) and (4)

$\Rightarrow 2g-4f+3=0\dots \left(6\right)$

Solving equation (5) and (6)

$g=\frac{-13}{2},f=\frac{-5}{2}$

From equation (4)

$\Rightarrow 2\times \frac{-13}{2}+2\times \frac{-5}{2}+c+2=0$

$\Rightarrow c=16$

Now put all three values in equation (1)

$x^2+y^2+2\times (-\frac{13}{2})\times x+2\times (-\frac{5}{2})\times y+16=0$

$\Rightarrow x^2+y^2-13x-5y+16=0$

Final answer:

$\therefore x^2+y^2-13x-5y+16=0$