If one end of a diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ be (3, 4), then the other end is

(0, 0)

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(1, 1)

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(1, 2)

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(2, 1)

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Solution

The correct option is C (1, 2)
Centre is (2, 3). One end is (3, 4).
$P_{2}$ divides the join of $P_{1}$ and O in ratio of 2 : 1.
Hence $P_{2} i s (\frac{4 - 3}{2 - 1}, \frac{6 - 4}{2 - 1}) \equiv (1, 2) .$

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