# Touching circles

GCSEs are over and the holiday is just kicking off. Time to return to WordPress with some trivial and not-so-trivial lemmata!
Here is what I discovered while doing some maths. I will give two proofs of this result – one using angles, and another using lengths (invoking some more advanced technology). Which do you prefer?

Circle A is internally tangent to larger circle B at the point Y. Points X and Z lie on the circle B such that the line XZ is tangent to circle A at the point M. Then MY bisects angle XYZ.

Proof 1
Let Obe the centre of A and let OB be the centre of B.

Observe that, since Y is the point of tangency between A and B, OB, OA and Y are collinear.
Let ∠MYO= α and let ∠OBYZ = β. We want to show that ∠XYM = α + β.
OAM = OAY (equal radii) so triangle OAMY is isosceles; therefore, ∠OAMY = α
Tangent and radius are perpendicular so ∠XMY = 90 – α
Also, OBY = OBZ (equal radii) so triangle OBYZ is isosceles; therefore, ∠OBZY = β
Since angles in a triangle sum to 180, ∠ZOBY = 180 – 2β
∠MXY = ½∠ZOBY = 90 – β (angle at centre is twice angle at circumference)
By angles in a triangle sum to 180, ∠XYM = 180 – ∠XMY – ∠MXY = α + β as required
Hence MY bisects angle XYZ.

Proof 2
Let XY and ZY cut the circle A at X’ and Z’ respectively, as shown:

Since the circles A and B touch at Y, we can consider Y as the centre of the enlargement that transforms circle A into circle B. Let the enlargement have scale factor K; then
$\frac{XY}{X'Y}=\frac{ZY}{Z'Y}=K$
Or
$\frac{X'Y}{XY}=\frac{Z'Y}{ZY}$
$1-\frac{X'Y}{XY}=\frac{XY-X'Y}{XY}=\frac{XX'}{XY}$
And
$1-\frac{Z'Y}{ZY}=\frac{ZY-Z'Y}{ZY}=\frac{ZZ'}{ZY}$
So
$\frac{XX'}{XY}=\frac{ZZ'}{ZY}$
which rearranges to
$\frac{XX'}{ZZ'}=\frac{XY}{ZY}$
Also, by power of a point/tangent-secant theorem/whatever you call it:
$MX^{2}=XX'\cdot XY$
And
$MZ^{2}=ZZ'\cdot ZY$
Dividing the first equation by the second,
$\frac{MX^{2}}{MZ^{2}}=\frac{XX'}{ZZ'}\frac{XY}{ZY}=(\frac{XY}{ZY})^{2}$
Square rooting:
$\frac{MX}{MZ}=\frac{XY}{ZY}$
Then, by the converse of the angle bisector theorem, MX bisects angle XYZ.

We can see that X’Z’ looks parallel to XZ. This can also be proved in two ways – firstly, by angles, using the above result in conjunction with alternate segment theorem – or, much more simply, by using a homothety/enlargement argument.