A circle is tangent to the $y$-axis at $(0,2)$, and the larger of its $x$-intercepts is 8 . The radius of the circle is
(A) $\frac{9}{2}$
(B) $\sqrt{17}$
(C) $\frac{17}{4}$
(D) $\frac{15}{4}$
(E) $\frac{\sqrt{17}}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-134.jpg?height=382&width=491&top_left_y=1717&top_left_x=1294)
A circle is tangent to the $y$-axis at $(0,2)$, and the larger of its $x$-intercepts is 8 . The radius of the circle is
(A) $\frac{9}{2}$
(B) $\sqrt{17}$
(C) $\frac{17}{4}$
(D) $\frac{15}{4}$
(E) $\frac{\sqrt{17}}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-263.jpg?height=423&width=563&top_left_y=274&top_left_x=1342)

Solution

Let the centre be $C$ and the radius $r$.

Then $C Q$ is perpendicular to the $y$-axis, and has length $r$.

So the $x$-coordinate of $C$ is $r$ and the $y$-coordinate is 2 , ie. $C(r, 2)$.

Drop a perpendicular $C S$ to the $x$-axis and consider right triangle CSP.

$C P=r, C S=2$ and $S P=8-r$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-263.jpg?height=417&width=575&top_left_y=797&top_left_x=1336)

By Pythagoras, $r^{2}=(8-r)^{2}+2^{2} \Rightarrow r^{2}=64-16 r+r^{2}+4 \Rightarrow 16 r=68 \Rightarrow r=\frac{17}{4}$.

ANSWER: (C)