Two circles have equations

x^2 + y^2 + 2ax + 2by + c = 0

and

x^2 + y^2 + 2a'x + 2b'y + c' = 0

Show that these circles are orthogonal if 2aa' + 2bb' = c + c'

I'm assuming you have to grad both of these to get an expression for the vector normal to the tangent at any point on each circle. However I am unsure of how to finish - I thought of dot producting them and setting this to zero, but would this be correct, as this would mean the radii/tangents are perpendicular no matter which points they were taken?

2. Originally Posted by joker_900
Two circles have equations

x^2 + y^2 + 2ax + 2by + c = 0

and

x^2 + y^2 + 2a'x + 2b'y + c' = 0

Show that these circles are orthogonal if 2aa' + 2bb' = c + c'

...
Hello,

1. The 2 circles must intersect therefore $|\overrightarrow{M_1 M_2}|< r_1 + r_2$..........That means you have to calculate first the radii and the distance between the 2 centers.

2. The tangent of the first circle must pass through the center of the second circle and the tangent of the second circle must pass threough the center of the first circle to get perpendicular tangents. (see attachment)

3. Thus $(r_1)^2 + (r_2)^2 = (|\overrightarrow{M_1 M_2}|)^2$ ......... After a few steps of transformation you should get the given result.

4. Good luck!