Let the two circles `S equiv x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0`

and `S' = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0`

intersect each other at the points `P` and `Q`. The angle `theta`
between two circles `S = 0` and `S' = 0` is defined as the
angle between the tangents to the two circles at the points
of intersection.

`c_1 , c_2` and `r_1, r_2` are centres and radius of circles `S = 0` and `S' = 0` then

`C_1 = (- g_1, - f_1)` and `C_1 = (- g_2, - f_2)` and

`r_1= sqrt((g_1^2 + f_1^2 - c_1 ))` and `r_2= sqrt( (g_2^2+f_2^2-c_2))`

Let `d= |C_1C_2|=` Distance between their centres

`=sqrt((-g_1+g_2)^2+(-f_1+f_2)^2)`

`=sqrt((g_1^2+f_1^2+g_2^2+f_2^2-2g_1g_2-2f_1f_2))`

Now in `Delta C_1PC_2, ` here `angle C_1PC_2=alpha`

`cos alpha=((r_1^2+r_2^2-d^2)/(2r_1r_2))`........(i) `(∵ alpha + theta+90^(circ) + 90^(circ)= 360^(circ))`

or `cos (180^(circ)-theta)=((r_1^2+r_2^2-d^2)/(2r_1r_2))` `(∵ alpha= 180^(circ) - theta)`

If the angle between the circles is `90^(circ)`, i.e., `theta = 90^(circ)`, then the circles are said to be orthogonal circles or we say that the circles cut each other ORTHOGONALLY.

Then from (i) ` quad quad quad quad quad quad quad 0=(r_1^2+r_2^2-d^2)/(2r_1r_2)`

or `r_1^2+r_2^2-d^2=0`

or `r_1^2+r_2^2-d^2=0`

or `r_1^2+r_2^2=d^2`

or Condition is orthogonality `2g_1g_2+2f_1f_2=c_1+c_2`