Question

Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x - 7y = 0 and whose centre is the point of intersection of the lines x +y +1 = 0 and x - 2y + 4 = 0.

Answer 1

The given equations of lines are
$x+3y=0\dots \left(1\right)2x-7y=0\dots \left(2\right)x+y-1\dots \left(3\right)x-2y=-4\dots \left(1\right)$
The general equation of circle with centre (a, b) and radius r is
$(x-a{)}^2+(y-b{)}^2=r^2\dots \left(A\right)$
centre of (A) is the point of intersection of (iii) and (iv)
$\therefore$ centre = (-2, 1)
$\therefore$ (A)
$\Rightarrow (x+2{)}^2+(y-1{)}^2=r^2\dots \left(B\right)$
Also, (A) passes through point of intersection of (1) and (2), that is through P = (0, 0)
$\therefore 2^2+(-1{)}^2=r^2\Rightarrow =r=\sqrt{5}$
Thus, the equation of required circle is
$(x+2{)}^2+(y-1{)}^2=5$
or $x^2+y^2+4x-2y=0$