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(A) Let $P$ be the center of circle $S$ and $r$ be its radius. Since it touches the $X$-axis at $A$ and $Y$-axis at $B$,the coordinates of $P$ are $(r

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$\frac{5\pi}{8}$

Vedclass · Answer

(A) Let $P$ be the center of circle $S$ and $r$ be its radius. Since it touches the $X$-axis at $A$ and $Y$-axis at $B$,the coordinates of $P$ are $(r, r)$.The distance from the origin $O(0,0)$ to $P(r,r)$ is $OP = \sqrt{r^2+r^2} = r\sqrt{2}$.Since the circle $S$ touches the unit circle $x^2+y^2=1$ externally at $C$,the distance between their centers is the sum of their radii: $OP = 1 + r$.Equating the two expressions for $OP$: $r\sqrt{2} = 1 + r \Rightarrow r(\sqrt{2}-1) = 1 \Rightarrow r = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1$.In $	riangle OAP$,$OA = r$,$AP = r$,and $\angle OAP = 90^{\circ}$. Thus,$	riangle OAP$ is an isosceles right triangle,so $\angle AOP = \angle APO = 45^{\circ}$.In $	riangle PCA$,$PC = r$ and $AC = r$ (radii of circle $S$). Thus,$	riangle PCA$ is an isosceles triangle with $\angle PCA = \angle PAC$.The angle $\angle CPA = 180^{\circ} - 45^{\circ} = 135^{\circ}$ (since $O, C, P$ are collinear).In $	riangle PCA$,$2\angle PCA + 135^{\circ} = 180^{\circ} \Rightarrow 2\angle PCA = 45^{\circ} \Rightarrow \angle PCA = 22.5^{\circ} = \frac{\pi}{8}$.Finally,$\angle OCA = 180^{\circ} - \angle PCA = 180^{\circ} - 22.5^{\circ} = 157.5^{\circ} = \frac{5\pi}{8}$.

Let $S$ be the circle in the $XY$-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 $\angle OCA$ equals:

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