The radius of the circle in which the sphere $x^{2} + y^{2} + \mathcal{z}^{2} + 2\mathcal{z} - 2y - 4\mathcal{z} - 19 = 0$ is cut by the plane $x + 2y + 2\mathcal{z} + 7 = 0$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Centre of the sphere is $( - 1,\ 1,\ 2)$ and its radius $= \sqrt{1 + 1 + 4 + 19} = 5$
$CL$, perpendicular distance of $C$ from plane, is $\left| \frac{- 1 + 2 + 4 + 7}{\sqrt{1 + 4 + 4}} \right| = 4$
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Now $AL^{2} = CA^{2} - CL^{2} = 25 - 16 = 9$
Hence, radius of the circle $= \sqrt{9} = 3$

Q. The radius of the circle in which the sphere $x^{2} + y^{2} + \mathcal{z}^{2} + 2\mathcal{z} - 2y - 4\mathcal{z} - 19 = 0$ is cut by the plane $x + 2y + 2\mathcal{z} + 7 = 0$ is

Explanation

Reviewed by Dr. Abubacker

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