Question:
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.
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).