# If one end of a d

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).

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