Question

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

• A. (0, 0)
• B. (1, 1)
• C. (1, 2)
• D. (2, 1)

Answer 1

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 $P_{2}is(\frac{4-3}{2-1},\frac{6-4}{2-1})\equiv (1,2).$