First slide

Planes in 3D

Question

The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 z - 2 y - 4 z - 19 = 0$ is cut by the plane x+2y+22+7=0 is

Moderate

A

2
B

3
C

1
D

4

Solution

Center of the sphere is (-1, 1,2) and its radius = $\sqrt{1 + 1 + 4 + 19} = 5$
C L, perpendicular distance of C from plane, is $|\frac{- 1 + 2 + 4 + 7}{\sqrt{1 + 4 + 4}}| = 4$
Now AL² = CA² - CL² =25 - 16=9
Hence, radius of the circle $\sqrt{9} = 3$

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