Question

The circle of equation $(x-3{)}^2+(y-2{)}^2=1$ has center c. Point $M(4,2)$ is on the circle. N is another point on the circle so that angle $McN$ has a size of ${30}^{\circ }$. Find the coordinates of point N.

• A. $(3+\sqrt{3/2},5/2)$
• B. $(5/2,3+\sqrt{3/2})$
• C. $(3-\sqrt{3/2},3/2)$
• D. $(3/2,3-\sqrt{3/2})$
• E. $(4,3)$

Answer 1

The correct option is A $(3+\sqrt{3/2},5/2)$

$(x-3{)}^2+(y-2{)}^2=1$

$C=(3,2),r=1$

Any point on the circle is given by

$(3+\cos \theta ,2+\sin \theta )$

Let N= $(3+\cos {\theta }_1,2+\sin {\theta }_1)$

Given $\angle MCN=30$

$⟹\tan \varphi =\frac{|m1–m2|}{1+m1m2}$

m1 and m2 are slopes of lines $\overline{MC}$ and $\overline{CN}$

$M_{\overline{CN}}=\frac{2+\sin {\theta }_1-2}{3+\cos {\theta }_1-3}=\tan \theta$

$M_{\overline{NC}}=\frac{2-2}{4-3}=0$

$\tan \varphi =\frac{\tan {\theta }_1-0}{1+0}=\tan 30$

$⟹{\theta }_1=30$

$⟹N=(3+\frac{\sqrt{3}}{2},2+\frac{1}{2})=(3+\frac{\sqrt{3}}{2},\frac{5}{2})$