Condition for Orthogonal Circles

Two circles x²+y²+2g₁x+2f₁y+c₁=0 and x²+y²+2g₂x+2f₂y+c₂=0 intersect orthogonally iff this condition holds. Equivalently d² = r₁²+r₂².

Two circles intersect orthogonally at a point $P$ if their tangents at $P$ are perpendicular, equivalently if their radii $O_1P$ and $O_2P$ are perpendicular.

In right triangle $O_1PO_2$ (right angle at $P$):

$$O_1O_2^2 = O_1P^2 + O_2P^2 \implies d^2 = r_1^2 + r_2^2$$

In terms of general form: Using $d^2 = (g_1-g_2)^2 + (f_1-f_2)^2$ and $r_i^2 = g_i^2 + f_i^2 - c_i$:

$$(g_1-g_2)^2 + (f_1-f_2)^2 = (g_1^2+f_1^2-c_1) + (g_2^2+f_2^2-c_2)$$

Expanding the left side:

$$g_1^2 - 2g_1g_2 + g_2^2 + f_1^2 - 2f_1f_2 + f_2^2 = g_1^2+f_1^2-c_1+g_2^2+f_2^2-c_2$$

Cancelling $g_1^2 + g_2^2 + f_1^2 + f_2^2$ from both sides:

$$-2g_1g_2 - 2f_1f_2 = -c_1 - c_2$$ $$\boxed{2g_1g_2 + 2f_1f_2 = c_1 + c_2}$$

Geometric property: Each circle passes through the centre of the other (not generally) — but the tangent from the centre of one circle to the other has length equal to the radius of the other. This is a useful alternative characterisation.