If one end of a diameter of the circle

Solution:

Given equation of the circle,

x2 – 4x + y2 – 6y + 11 = 0

x2 – 4x + 4 + y2 – 6y + 9 +11 – 13 = 0

the above equation can be written as

x2 – 2 (2) x + 22 + y2 – 2 (3) y + 32 +11 – 13 = 0

on simplifying we get

(x – 2)2 + (y – 3)2 = 2

(x – 2)2 + (y – 3)2 = (√2)2

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)2 + (y – k)2 = r2

We have centre = (2, 3)

The centre point is the mid-point of the two ends of the diameter of a circle.

Let the points be (p, q)

(p + 3)/2 = 2 and (q + 4)/2 = 3

p + 3 = 4 & q + 4 = 6

p = 1 & q = 2

Hence, the other ends of the diameter are (1, 2).