## Question

### Solution

Correct option is The sphere is .     Centre = (–1, 1, 2) and radius = 5.

The plane is CA = length of perpendicular from C(–1, 1, 2) to the plane = 4 < radius of sphere

∴ Given plane intersects the sphere. ∴ Radius of circle with centre at A = 3.

AC is normal to the given plane.

∴ D.r.’s of AC are 1, 2, 2. AC passes through C(–1, 1, 2). Let the coordinates of A be ( Since, A lies on the plane, we have (     #### SIMILAR QUESTIONS

Q1

Find the equation of the sphere whose centre has the position vector and which touches the plane Q2

Find the value of for which the plane touches the sphere Q3

Find the radius of the circular section of the sphere by the plane Q4

Find the equation of the sphere passing through the points (0, 0, 0), (–1, 2, 0), (0, 1, –1) and (1, 2, 5).

Q5

Find the equation of the sphere which passes through the points (1,–3, 4), (1, –5, 2), (1, –3, 0) and has its centre on the plane x + y + z = 0.

Q6

Chord AB is a diameter of the sphere with coordinates of A as (3, 2, –2). Find the coordinates of B.

Q7

A plane passes through a fixed point (­a, b, c) and cuts axes in A, B, C. Find the locus of the centre of the sphere OABC