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.
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).