If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 =

1.	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.
Answer» 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).

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.

Answer»

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).