The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :

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Answer: D. (–3, 6) Let S be the circle passing through point of intersection of S1 & S2 	herefore S = S1 + \lambda S2 = 0 \Rightarrow S:({x^2} + {y^2} - 6x) + \lambda ({x^2} + {y^2} - 4y) = 0 \Rightarrow S:{x^2} + {y^2} - \left( {{6 \over {1 + \lambda }}} ight)x - \left( {{{4\lambda } \over {1 + \lambda }}} ight)y = 0 ....(1) Centre \left( {{3 \over {1 + \lambda }},{{2\lambda } \over {1 + \lambda }}} ight) lies on 2x - 3y + 12 = 0 \Rightarrow \lambda = - 3 put in (1) \Rightarrow S:{x^2} + {y^2} + 3x - 6y = 0 Now check options point ( - 3,6) lies on S.

The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :

Answer

Explanation

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