Question

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

Answer 1

We have, $2x-3y=-4\dots \left(1\right)3x+4y=5\dots \left(2\right)$
The point of ntersection of (1) and (2) is\
$P=(\frac{-1}{17},\frac{66}{51})orP=(\frac{-1}{17},\frac{22}{17})$
According to the equation, centre=$P=(\frac{-1}{17},\frac{22}{17})$
Also, the circle passes through 0(0,0)
$\therefore r=OC=\sqrt{{(0+\frac{1}{17})}^2+{(0-\frac{1}{17})}^2}=\sqrt{\frac{1}{289}+\frac{484}{289}}=\frac{\sqrt{485}}{17}$
Thus, the required equation of circle is
${(x+\frac{1}{17})}^2+{(y-\frac{22}{17})}^2=\frac{485}{289}$