Question

Let $S$ be the circle in $xy$-plane which touches the $x$-axis at point $A$, the $y$-axis at point $B$ and the unit circle $x^2+y^2=1$ at point C externally. If $O$ denotes the origin, then the angle $OCA$ equals

• A. $\frac{5\pi }{8}$
• B. $\frac{\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $\frac{3\pi }{5}$

Answer 1

The correct option is A $\frac{5\pi }{8}$


$AD=a=CD$ and $OC=1$
Using pythagoras in $△OBD$,
$O{B}^2+B{D}^2=O{D}^2\Rightarrow OD=a\sqrt{2}$
Now from the figure we can write,
$OC+CD=OD\Rightarrow 1+a=a\sqrt{2}\Rightarrow (1+a{)}^2=2a^2\Rightarrow 1+a^2+2a=2a^2\Rightarrow a^2-2a-1=0\Rightarrow a=1+\sqrt{2}$
[$\because a>1$]
Coordinates of $C=(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$
Cordinates of $A=(-(1+\sqrt{2}),0)$
Length of $AC=\sqrt{2+\sqrt{2}}$
Using cosine rule in the $△OCA$
Let $\angle OCA=\theta$
$\cos \theta =\frac{A{C}^2+O{C}^2-O{A}^2}{2(OC\times AC)}=\frac{2+\sqrt{2}+1-(1+\sqrt{2}{)}^2}{2\sqrt{(2+\sqrt{2})}}=\frac{-\sqrt{2}}{2\sqrt{(2+\sqrt{2})}}=\frac{-1}{\sqrt{2}}\times \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2}}$
As $\theta$ is an angle of triangle so $\theta <\pi$
and as $\cos \theta$ is negative,
$\frac{\pi }{2}<\theta <\pi$
$\therefore 2{\cos }^2\theta =\frac{2-\sqrt{2}}{2}=1+\cos 2\theta \Rightarrow \cos 2\theta =-\frac{1}{\sqrt{2}}$
So $2\theta =\frac{3\pi }{4}\text{or}\frac{5\pi }{4}\therefore \theta =\frac{5\pi }{8}$
[$\because \frac{\pi }{2}<\theta <\pi$]
Hence the angle $OCA=\frac{5\pi }{8}$