Answer: 6
SOLUTION
∵segment[tex]\frac{}{AB}[/tex]is tangent to the circle at point C.
∴∠MCA=∠MCB=90°
Suppose [tex]\frac{}{MC} =x[/tex]
From the Pythagorean theorem
MA²=9²+[tex]x[/tex]²
MB²=4²+[tex]x[/tex]²
∴MA²+MB²=AB²
i.e. [tex]x^{2} +9^{2} +4^{2} +x^{2} =(9+4)^{2} \\2x^{2} +97=169\\2x^{2} =72\\x^{2} =36\\x=6[/tex]
We conclude that [tex]x=6=\frac{}{MC}[/tex]
