Question:
The circle passing through the intersection of the circles,
\(x^2 + y^2 - 6x = 0\) and \(x^2 + y^2 - 4y = 0\), having its centre on
the line, \(2x - 3y + 12 = 0\), also passes through the point :
The circle passing through the intersection of the circles,
\(x^2 + y^2 - 6x = 0\) and \(x^2 + y^2 - 4y = 0\), having its centre on
the line, \(2x - 3y + 12 = 0\), also passes through the point :
\(x^2 + y^2 - 6x = 0\) and \(x^2 + y^2 - 4y = 0\), having its centre on
the line, \(2x - 3y + 12 = 0\), also passes through the point :
Show Hint
Use the family \(S_1 + \lambda S_2 = 0\), find its centre in terms of \(\lambda\), put it on the given line, then test the options.
Updated On: Nov 5, 2026
- \((-3, 1)\)
- \((1, -3)\)
- \((-1, 3)\)
- \((-3, 6)\)
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The Correct Option is D
Solution and Explanation
Family of circles through the intersection: \(x^2 + y^2 - 6x + \lambda(x^2 + y^2 - 4y) = 0\), i.e. \((1+\lambda)(x^2+y^2) - 6x - 4\lambda y = 0\).
Centre: \(\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)\).
It lies on \(2x - 3y + 12 = 0\): \(\frac{6}{1+\lambda} - \frac{6\lambda}{1+\lambda} + 12 = 0 \Rightarrow 6 - 6\lambda + 12 + 12\lambda = 0 \Rightarrow \lambda = -3\).
Circle: \(-2(x^2+y^2) - 6x + 12y = 0 \Rightarrow x^2 + y^2 + 3x - 6y = 0\).
Test \((-3, 6)\): \(9 + 36 - 9 - 36 = 0\), satisfied. (Others: \((-3,1)\): \(9+1-9-6 = -5\); \((1,-3)\): \(1+9+3+18 = 31\); \((-1,3)\): \(1+9-3-18 = -11\); none zero.)
Answer: (4) (-3, 6)
Centre: \(\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)\).
It lies on \(2x - 3y + 12 = 0\): \(\frac{6}{1+\lambda} - \frac{6\lambda}{1+\lambda} + 12 = 0 \Rightarrow 6 - 6\lambda + 12 + 12\lambda = 0 \Rightarrow \lambda = -3\).
Circle: \(-2(x^2+y^2) - 6x + 12y = 0 \Rightarrow x^2 + y^2 + 3x - 6y = 0\).
Test \((-3, 6)\): \(9 + 36 - 9 - 36 = 0\), satisfied. (Others: \((-3,1)\): \(9+1-9-6 = -5\); \((1,-3)\): \(1+9+3+18 = 31\); \((-1,3)\): \(1+9-3-18 = -11\); none zero.)
Answer: (4) (-3, 6)
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