Question 1000937
Hi there,
With equation of a circle:
x^2 + y^2 + 6x - 4y - 87 = 0
The centre of the circle is (-3,2)
The radius = 10
(sqrt (-3)^2 + (2)^2 -(-87))
If you move along horizontally from 
coordinates (-3,2) by 10 units the 
set of coordinates (7,2) is reached.
This point is on the circle.
These will satisfy the equation:-
x^2 + y^2 + 6x - 4y - 87 =0
(7)^2 +(2)^2 + 6(7) - 4(2) - 87 = 0
49 + 4 + 42 - 8 - 87 = 0
Hope this helps :-)