The equation of the circle passing through the point (4, 6) and whose diameters are along x + 2y − 5 = 0 and 3x − y − 1 = 0 is
The centre is where the two diameters meet: \(x + 2y = 5\) and \(3x - y = 1\) give \(x = 1\), \(y = 2\).
Radius² = \((4-1)^2 + (6-2)^2 = 9 + 16 = 25\)
Circle: \((x-1)^2 + (y-2)^2 = 25 \Rightarrow x^2 + y^2 - 2x - 4y - 20 = 0\)
Answer: (C)
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