Question

Find the equation of the circle whose centre is the point of intersection of the lines $2x-3y+4=0$ & $3x+4y-5=0$ and passes through the origin.

Answer 1

$2x-3y+4=0$

$3x+4y-5=0$

To find eqn of circle passing through $(0,0)$ & centre at point of intersection

$\Rightarrow 2x-3y+4=0$ ........ $\left(i\right)$

$3x+4y-5=0$ ........... $\left(ii\right)$

Multiply $3\times \left(i\right)$ & $2\times \left(ii\right)$ & Subtract them, now we get.

$2(3x+4y-5)=0$

$3(2x-3y+4)=0$

$6x+18-10=0$

$-6x-9+12=0$

______________

$17y-22=0$

$\therefore =\frac{22}{17}$

on substituting $y=\frac{22}{17}$in $\left(i\right)$ we get

$2x=3y-4$

$=3\left(\frac{22}{17}\right)-4=\frac{66}{17}-68$

$\therefore x=\frac{-1}{17}$

$\therefore$ centre of circle is at $(\frac{-1}{17},\frac{22}{17})$

Since radius of circle $=$ Distance from centre & origin

$\therefore r=\sqrt{{(-\frac{1}{17}-0)}^2+(\frac{22}{17}-0)}$

$r=\sqrt{\frac{1}{289}+\frac{484}{289}}$

$r^2=\frac{485}{289}$

eqn of circle $=(x-h{)}^2+(y-k{)}^2=r^2$

on substituting all values we get

$\Rightarrow {(x+\frac{1}{17})}^2+{(y-\frac{22}{17})}^2=\frac{485}{289}$