The radius of the circle of intersection of the sphere $x^{2} + y^{2} + z^{2} = 9$ by the plane $3 x + 4 y + 5 z = 5$

\sqrt{\frac{17}{2}}

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3

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\sqrt{34}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is A $\sqrt{\frac{17}{2}}$
Centre and radius of the given sphere are $(0, 0, 0)$ and $3$ respectively.
Now the perpendicular distance between centre of sphere to the given plane is given by,
$d =∣ \frac{- 5}{\sqrt{3^{2} + 4^{2} + 5^{2}}} ∣= 1 / \sqrt{2}$
Therefore radius of the circle in which the given sphere and plane cut is
$= \sqrt{3^{2} - (1 / \sqrt{2})^{2}} = \sqrt{\frac{17}{2}}$

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