Question

The plane 2x + 2y – z = k touches the sphere

x² + y² + z² – 4x + 2y – 6z + 5 = 0 and makes a positive intercept on the axis of z, then k =

• A. –10
• B. –8
• C. 8
• D. 10

Answer 1

Answer: (A)

–10

Answer 2

Centre of the given sphere is (2, –1, 3) and is radius is

$\sqrt{4 + 1 + 9 - 5}$= 3.

As the given plane touches the given sphere

$\frac{2 \times 2 + 2( - 1) - (3) - k}{\sqrt{2^{2} + 2^{2} + ( - 1)^{2}}}$ = ±3

Ž k = 8 or k = –10

Since the plane makes a positive intercept on z-axis k < 0 and the required value of k = –10.