Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 6x – 14y – 1 = 0 and x2 + y2 – 4x + 10y – 2 = 0.

Given that we need to find the equation of the circle whose end points of a diameter are the centres of the circles



x2 + y2 + 6x - 14y - 1 = 0 .... - (i) and


x2 + y2 - 4x + 10y - 2 = 0 ...... - (ii)


Let us assume A and B are the centres of the 1st and 2nd circle.


We know that for a circle x2 + y2 + 2ax + 2by + c = 0 ...... - (1)


Centre = (- a, - b)


Radius =


Comparing (i) with (1) we get,


Centre(A) =


A = (- 3,7)


Comparing (ii) with (1) we get,


Centre(B) =


B = (2, - 5)


We know that the centre is the mid - point of the diameter.


Centre(C) =



We have a circle with centre and passing through the point (2, - 5).


We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.


We know that the distance between the two points (x1,y1) and (x2,y2) is .


Let us assume r is the radius of the circle.







We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:


(x - p)2 + (y - q)2 = r2


Now we substitute the corresponding values in the equation:




x2 + y2 + x - 2y - 41 = 0


The equation of the circle is x2 + y2 + x - 2y - 41 = 0.


2