Question

In the figure, segment $MN$ is a chord of a circle with center $O$, $MN=25$, $L$ is a point on chord $MN$ such that $ML=9$ and $d(O,L)=5$. Find the radius of the circle.

Answer 1

From $\Delta AOB$

$\Rightarrow \frac{25}{2}=9+5\cos \theta$

$\Rightarrow \frac{25}{2}-9=5\cos \theta$

$\Rightarrow 5\cos \theta =\frac{25-18}{2}=\frac{7}{2}$

$\therefore \cos \theta =\frac{7}{10}$

$\Rightarrow 5\sin \theta =5\sqrt{1-{\cos }^2\theta }$

$=5\sqrt{1-\frac{49}{100}}=5\sqrt{\frac{100-49}{100}}$

$=5\sqrt{\frac{51}{100}}$

$=\frac{5}{10}\sqrt{51}=\frac{\sqrt{51}}{2}$

$\therefore 5\sin \theta =\frac{\sqrt{51}}{2}$

$r=\sqrt{{\left(\frac{25}{2}\right)}^2+{\left(\frac{\sqrt{51}}{2}\right)}^2}$

$=\frac{1}{2}\sqrt{625+51}=\frac{1}{2}\sqrt{676}$

$=\sqrt{\frac{676}{4}}$

$=\sqrt{169}=13$.