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).

∴ \[\left( 0 + 2 \right)^2 + \left( 0 - 1 \right)^2 = a^2\]

\[\Rightarrow 4 + 1 = a^2 \]
\[ \Rightarrow a = \sqrt{5} \left( \because a > 0 \right)\]

Substituting the value of a in equation (1):

\[\left( x + 2 \right)^2 + \left( y - 1 \right)^2 = 5\]

\[\Rightarrow x^2 + 4 + 4x + y^2 + 1 - 2y = 5\]
\[ \Rightarrow x^2 + 4x + y^2 - 2y = 0\]

Hence, the required equation of the circle is

\[x^2 + y^2 + 4x - 2y = 0\]