geometry
A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.
There are two tangents to the circle which pass through point $P$ – one on each side. They intersect the circle at points $A$ and $B$.
What is the angle through $P$ between these two tangents? In other words, angle $APB$?
I know that angle $APB$ + angle $ACB$ add up to 180.
(Not homework, for graphics programming)
Thanks,
Louise
Hint: Try to use congruence. If you can prove that the two triangles are congruent then it is a kite. Now, you can prove it easily.
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).
Best Answer
Here is a picture:
$\overline {CP} = d$
$\angle CAP$ and $\angle BAP$ are right angles, and $\triangle APB$ is isosceles.
$m\angle APC = \arcsin \frac rd\\ m\angle APB = 2\arcsin \frac rd\\ m\angle BAP = \arccos \frac rd$