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 point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 is (0, 0).

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

The point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0 is (−2, 1).

∴ h = −2, k = 1

∴ Equation of the required circle = ${\left(x+2\right)}^2+{\left(y-1\right)}^2=a^2$ ...(1)

Also, equation (1) passes through (0, 0).

∴ ${\left(0+2\right)}^2+{\left(0-1\right)}^2=a^2$
$$\begin{gathered} \Rightarrow 4+1=a^2 \\ \Rightarrow a=\sqrt{5}\left(\because a>0\right) \end{gathered}$$

Substituting the value of a in equation (1):
${\left(x+2\right)}^2+{\left(y-1\right)}^2=5$
$$\begin{gathered} \Rightarrow x^2+4+4x+y^2+1-2y=5 \\ \Rightarrow x^2+4x+y^2-2y=0 \end{gathered}$$

Hence, the required equation of the circle is $x^2+y^2+4x-2y=0$.