So I'm stumped by what should be a rather simple problem. There are two circles whose tangents intersect at each others' centres. The tangents are at right angle. If I know the distance between the centres, there should be simple geometry to solve the radii of the circles.
I know I could do like an equation group of this; call radius one $a$, radius two $b$, write down Pythagoran theorem for the triangle in the middle, then maybe like trigonometric functions for the halves of the two central angles of the circles, but I can't believe it should be this complex (that would be what, a four equation group?). There's something about the symmetry of these circles I'm missing that ought to make this simpler.
I know I have:
Let the radius of circle $A$ be $a$, and circle $B$ be $b$. Further, let the line drawn at their intersection be $c$, and the line from $c$ to centre of circle $B$ be $d$.
$$a^2 + b^2 = 10^2 \\a^2 + (10-d)^2 = \left( \frac c 2 \right) ^2 \\
b^2 + d^2 = \left( \frac c 2 \right) ^2 $$
If I add to that trigonometric functions and the sum of the angles, I believe I could solve them but it just feels way too complex and certainly not the natural solution.
An aside: does anyone know how these circles (ones whose tangents drawn at the intersection points of the circles are the centres of one another) are called? I know there's a term for it but I can't find it for the life of me.
Best Answer
They are called orthogonal circles.
The necessary and sufficient condition for two circles to be orthogonal can be given under two different forms :
$$R^2+R'^2=d^2 \ \ \ \ \iff \ \ \ \ aa'+bb'=\tfrac12(c+c')\tag{1}$$
where $R,R'$ are their radii,
$$d=\sqrt{(a-a')^2+(b-b')^2} \ \ \text{ the distance between their centers}$$ and
$$\begin{cases}x^2+y^2-2ax-2by+c&=&0\\ x^2+y^2-2a'x-2b'y+c'&=&0\end{cases} \ \ \ \text{their cartesian equations}$$
There is a very nice representation of (1) in the so-called "space of circles", where a circle with equations
$$x^2+y^2-2ax-2by+c=0 \ \ \ \iff \ \ \ (x-a)^2+(y-b)^2=R^2\tag{2}$$
is represented by 3 coordinates, $(a,b,c)$.
Please note the relationship :
$$a^2+b^2-R^2=c\tag{3}$$
If we write (3) under the form :
$$R^2=\underbrace{a^2+b^2-c}_{\|\sigma\|^2}\tag{3}$$
it gives us the opportunity to define the norm of a circle (nothing scandalous : the norm of a circle is plainly its radius).
We now define the dot product between 2 circles with coordinates $(a,b,c)$ and $(a',b',c')$ by :
$$\sigma \ \cdot \ \sigma' \ := \ aa'+bb'-\dfrac12(c+c'). \tag{4}$$
One can show easily the nice relationship:
$$\sigma \ \cdot \ \sigma' \ = \|\sigma\|\|\sigma'\| \cos \alpha$$
(proof below)
with $\alpha$ defined as the angle between the radii at intersection point $I$.
Particular case : is $\alpha =\dfrac{\pi}{2}$, we find relationship (1) !
Fig. 1 : 2 non-orthogonal circles illustrating the notations for relationship (4).
Appendix : Proof of relationship (4).
The law of cosines in triangle $OIO'$ gives:
$$d^2=R^2+R'^2-2RR' \cos \alpha$$
which can be written :
$$(a-a')^2+(b-b')^2=a^2+b^2-c+a'^2+b^2-c-2RR' \cos \alpha$$
Expanding the LHS and simplifying, we get indeed :
$$2aa'+2bb'-(c+c')=2RR' \cos \alpha$$
which nothing else than (4).