Question

Find the equation of the circle which passes through the point of intersection of the lines $3x-2y-1=0$ and $4x+y-27=0$ and whose centre is $(2,-3)$

Answer 1

Let P be the point of intersection of the lines AB and LM whose equations are respectively
$3x-2y-1=0$ .......(i)
and $4x+y-27=0$ .......(2)
From equation(i)$+2\times$equation(2), we get
$\Rightarrow 3x-2y-1+8x+2y-54=0$
$\Rightarrow 11x=55$
$\therefore x=5$
Substitute $x=5$ in equation (2), we get
$20+y-27=0$
$\Rightarrow y-7=0$
$\therefore y=7$
Thus, we get $x=5,y=7$
So coordinates of P are $(5,7)$
Let $C(2,-3)$ be the centre of the circle
Since the circle passes through P
Therefore CP = radius = $\sqrt{{(5-2)}^2+{(7+3)}^2}$
$=\sqrt{9+100}$
$\Rightarrow radius=\sqrt{109}$
Hence, the equation of the required circle is
${(x-2)}^2+{(y+3)}^2={\left(\sqrt{109}\right)}^2$