Orthogonal Circles

Definition. Two intersecting circles are called orthogonal if the angle between them (as defined in the previous section) is exactly $90^\circ$ — the two tangent lines at the intersection point are perpendicular to each other.

Why this is a natural special case. Since a radius is always perpendicular to its own circle's tangent, saying "the two tangents are perpendicular" is the same as saying "each circle's tangent at $P$ passes straight through the other circle's centre." Put differently: if you stand at the centre of one circle and draw a line to the point of intersection, that line is tangent to the other circle. This is a useful mental picture — orthogonality is really about each circle's radius lining up with the other circle's tangent direction.

Deriving the condition. Orthogonality means $\theta=90^\circ$, i.e. $\cos\theta=0$. Plugging into the formula from §2.2,

giving the clean, purely-algebraic condition for orthogonality:

Note there are no square roots left at all — you can check orthogonality by inspection of the six coefficients, without ever computing a centre, a radius, or an angle. Equivalently, in terms of centres/radii, orthogonality holds exactly when $d^2 = r_1^2+r_2^2$ (this drops straight out of the $\cos\theta=0$ version of the §2.2 formula, and is really just the Pythagorean theorem applied to $\triangle C_1PC_2$, which becomes right-angled at $P$).

Worked example. Find the value of $k$ for which the circles

cut each other orthogonally.

Solution. Here $g=-3,\ f=2,\ c=2$ and $g'=k,\ f'=-1,\ c'=-24$. Applying the orthogonality condition:

$-6k-4=-22 \ \Longrightarrow\ -6k=-18 \ \Longrightarrow\ k=3.$ …