If one end of a diameter of the circle x² + y² – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.

Equation of the circle,

x² - 4x + y² - 6y + 11 = 0

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

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

(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

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 = 4 & q + 4 = 6

p = 1 & q = 2

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