Question

Find the equation of circle which circumscribes the triangle forms by the lines $y=x+2,3y=4x$ and $2y=3x$

Answer 1

Let A is the Point of intersction of line $y=x+2and3y=4x$

$\Rightarrow y=\frac{4}{3}x$

Put value of $y$ in $y=x+2$

$\Rightarrow \frac{4}{3}x=x+2$

$\Rightarrow x=6$

$\because x=6\Rightarrow y=8$

$\therefore A(6,8)$

Origin is the Point of intersction of line $3y=4x$ and $2y=3x$

Let B is the Point of intersction of line $y=x+2and2y=3x$

$\Rightarrow y=\frac{3}{2}x$

Put value of $y$ in $y=x+2$

$\Rightarrow \frac{3}{2}=x+2$

$\Rightarrow x=4$

$\because x=4\Rightarrow y=6$

$\therefore B(4,6)$

Origin is the Point of intersction of line $3y=4x$ and $2y=3x$

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 $OAB$

And for point $O(0,0)$

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

$\Rightarrow c=0\dots \left(2\right)$

For point $A(6,8)$

$6^2+8^2+2g\times 6+2f\times 8+c=0$

$\Rightarrow 12g+16f+100=0$

$\Rightarrow 9g+12f+75=0\dots \left(3\right)$

For point $B(4,6)$

$4^2+6^2+2g\times 4+2f\times 6+c=0$

$\Rightarrow 8g+12f+52=0\dots \left(4\right)$

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

Subtract equation (3) and (4)

$\Rightarrow g+23=0$

$\Rightarrow g=-23$

From equation (4)

$\Rightarrow 8\times (-23)+12f+52=0$

$f=11$

Now put all three values in equation (1)

$x^2+y^2+2\times (-23)\times x+2\times 11\times y+0=0$

$\Rightarrow x^2+y^2-46x+22y=0$

Final answer:

$\therefore x^2+y^2-46x+22y=0$