A circle $\Gamma_2$ centered at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$, and the segments $PQ$ and $CD$ intersect at $M$

Question

Let $CD$ be a chord of circle $\Gamma_1$ and $AB$ be a diameter perpendicular to $CD$ at $N$ with $AN > NB$. A circle $\Gamma_2$ centered at $C$ with radius $CN$ intersects $\Gamma_1$ at points $P$ and $Q$, and the segments $PQ$ and $CD$ intersect at $M$. Given that the radii of $\Gamma_1$ and $\Gamma_2$ are $61$ and $60$ respectively, find $AM$.

What I Tried: Here is a picture:-

I couldn't understand the question properly, as it is written that $AB \perp CD$ but I cannot understand in what point the perpendicular line is drawn. The problem overall also seems complicated so I cannot understand anyway on how to start solving it.

Can anyone help me? Thank You.

Edit:- As @cosmos5 mentioned in the comments, my drawing as well my thinking was totally wrong. Here is the new picture :-

As usual, the problem seems complicated and I can't think of anyway to start this, I can however see that we might use Pythagorean Theorem in some form or the other but I cannot understand how since I don't know any lengths other than the radii of the circles.

Answer 1

M lies on common chord of the two circles. The common chord is locus of points having same power wrt the two circles.

Formula for power of a point $X$ is $OX^2-r^2$. So we have

$$MC^2-60^2=MO^2-61^2 $$

Substituting $MN=x$, we get $$(60-x)^2-60^2=x^2+11^2-61^2$$ $$\Rightarrow MN=30$$

$$\therefore AM=\sqrt{AN^2+MN^2}=78$$

Answer 2

$CN = 60$

$ON^2 = 61^2 - 60^2 \implies ON = 11$

$AN = AO + ON = 72$.

Now we need to find $MN$ and we then know $AM$.

$MN = CN - CM = 60 - CM$

If $\angle CON = \theta$, $\sin \theta = \frac{60}{61}$

$CM = \frac{CF}{\sin \theta}$ ($\angle FCM = 90^0 - \theta)$

$CF^2 = 60^2 - PF^2$

$OF^2 = (61-CF)^2 = 61^2 - PF^2$

Solving you get $CM = MN = 30$.

$AM = \sqrt{72^2 + 30^2} = 78$