Consider a circle C which touches the y-axis at (0,5) and cuts off an intercept $3 \sqrt{2}$ on the x-axis. Then the centre and the radius of the circle C are:

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$(5, \pm \frac{\sqrt{118}}{2}), \sqrt{\frac{59}{2}}$

$(\pm \frac{\sqrt{118}}{2}, 5), \sqrt{\frac{59}{2}}$

$(\pm \frac{\sqrt{118}}{2}, 5), 10$

$(\pm \frac{\sqrt{118}}{2}, 5), 5$

answer is D.

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Detailed Solution

Since the circle touches $y -$ axis at $(0, 5)$ , equation can be written as

$x^{2} + y^{2} - 2 a x - 10 y + 25 = 0$

Given that $x -$ Intercept is $3 \sqrt{2}$

it implies that

$2 \sqrt{g^{2} - c} = 3 \sqrt{2}$

Squaring on both sides and then substitute the appropriate values

$\begin{array}{rcl} 4 (g^{2} - c) & = & 18 \\ g^{2} - c & = & \frac{9}{2} \\ g^{2} & = & \frac{9}{2} + 25 \\ = & \frac{59}{2} \end{array}$

Hence, the circle is $x^{2} + y^{2} \pm \sqrt{118} x - 10 y = 25 = 0$

Hence, the center is $C (\mp \frac{\sqrt{118}}{2}, 5),$ and the radius is $r = \sqrt{\frac{118}{4} + 25 - 25} = \sqrt{\frac{59}{2}}$

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