Question

The equation of circle circumscribing the triangle whose sides are the line $y=x+2,3y=4xand2y=3x$ is

Answer 1

Let A is the point of intersection of line $y=x+2$ and $3y=4x$

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

Substituting value of y in $y=x+2,$

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

$\Rightarrow x=6$

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

$\Rightarrow A\equiv (6,8)$

Origin is the point of intersection of line $3y=4x$ and $2y=3x$

Ley B is the point of intersection of line $y=x+2$ and $2y=3x$

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

Substituting value of y in $y=x+2$,

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

$\Rightarrow x=4$

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

$\therefore B=(4,6)$

Origin is the point of intersection of line $3y=4x$ and $2y=3x$

Let equation of such a circle be

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

Since, circle passes through the vertices of triangle OAB and for point O(0,0)

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

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

For point A (6, 8),

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

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

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

For point B (4, 6),

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

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

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

Subtracting equation (3) and (4)

$\Rightarrow g+23=0$

$\Rightarrow g=-23$

From equation (4),

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

$\Rightarrow f=11$

Now substituting all three values in equation (1),

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

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

The equation of circle circumscribing the triangle whose sides are the line $y=x+2,3y=4x$ and $2y=3x$ is $x^2+y^2-46x+22y=0$