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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.

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Solution

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 x-h2+y-k2=a2.

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 = x+22+y-12=a2 ...(1)

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

0+22+0-12=a2
4+1=a2a=5 a>0

Substituting the value of a in equation (1):
x+22+y-12=5
x2+4+4x+y2+1-2y=5x2+4x+y2-2y=0

Hence, the required equation of the circle is x2+y2+4x-2y=0.

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