Question: If one end of the diameter of a circle x² + y² -4x -6y +12 = 0 is (1,3), find the co-ordinates of the other end.

Solution:
Given equation of a circle is x² + y² - 4x - 6y + 12 = 0

Let AB be the diameter of that circle. Coordinates of point A (x1,y1) = (1,3). So, let the coordinates of the point B be (x2,y2).

We have,
Equation of circle is;

x² + y² - 4x - 6y + 12 = 0

or, x² - 4x + 3 + y² - 6y + 9 = 0

or, (x² - x - 3x +3) + (y² - 2.3y + 3²) = 0

or, {x(x -1) -3(x -1)} + (y -3)² = 0

or, (x - 1)(x - 3) + (y - 3)(y -3) = 0 - (i)

Comparing equation (i) with the equation of circle (x - x1)(x - x2) + (y - y1)(y - y2) = 0, we get,

x2 = 3, y2 = 3

Hence, the required coordinate of the other end of the diameter is (3,3).

#SciPiPupil