5. Let $C D$ be a chord of a circle $\Gamma_{1}$ and $A B$ a diameter of $\Gamma_{1}$ perpendicular to $C D$ at $N$ with $A N>N B$. A circle $\Gamma_{2}$ centred at $C$ with radius $C N$ intersects $\Gamma_{1}$ at points $P$ and $Q$, and the segments $P Q$ and $C D$ intersect at $M$. Given that the radii of $\Gamma_{1}$ and $\Gamma_{2}$ are 61 and 60 respectively, find the length of $A M$.
5. Answer: 78
Extend $D C$ meeting $\Gamma_{2}$ at $H$. Note that $D N=N C=C H=60$. Since $M$ is of equal power with respect to $\Gamma_{1}$ and $\Gamma_{2}$. Thus $M N \cdot M H=M C \cdot M D$. That is $M N(M C+60)=$ $M C(M N+60)$ giving $M N=M C$. Thus $M N=30$.

The power of $N$ with respect to $\Gamma_{1}$ is $D N \cdot N C=60^{2}$, and is also equal to $N A \cdot N B=$ $N A \cdot(A B-N A)=N A \cdot(122-N A)$. Thus $N A \cdot(122-N A)=60^{2}$. Solving this quadratic equation, we get $N A=72$ or 50 . Since $N A>N B$, we have $N A=72$. Consequently $A M=\sqrt{N A^{2}+M N^{2}}=\sqrt{72^{2}+30^{2}}=78$