Equation of a sphere Practice Questions - NDA Mathematics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

If p is the perpendicular distance from the centre of the sphere to the plane, then which one of the following is correct?

Step-by-Step Logic & Solution

Question 2 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

The equation of the sphere S is

x² + y² + z² − 4x − 6y − 12z + k = 0.

What is the radius of the sphere passing through origin and concentric with the sphere S?

Step-by-Step Logic & Solution

Solution

Option (c) is correct.

Explanation:

The given equation of the sphere S is

\[ x^2 + y^2 + z^2 - 4x - 6y - 12z + k = 0 \]

To find the center and radius of the sphere, we complete the square for each variable. For \(x\): \[ x^2 - 4x \rightarrow (x-2)^2 - 4 \] For \(y\): \[ y^2 - 6y \rightarrow (y-3)^2 - 9 \] For \(z\): \[ z^2 - 12z \rightarrow (z-6)^2 - 36 \] Substituting back into the equation gives: \[ (x-2)^2 + (y-3)^2 + (z-6)^2 - 4 - 9 - 36 + k = 0 \] Thus, the center of the sphere \(C\) is \((2, 3, 6)\) and the equation can be rewritten as: \[ (x-2)^2 + (y-3)^2 + (z-6)^2 = 49 - k \] This reveals that the radius of the sphere is given by: \[ r = \sqrt{49 - k} \] For the sphere to pass through the origin \((0, 0, 0)\), we use the distance formula to find the distance from the center to the origin: \[ d = \sqrt{(2 - 0)^2 + (3 - 0)^2 + (6 - 0)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Thus, we set the radius equal to this distance: \[ \sqrt{49 - k} = 7 \] Squaring both sides yields: \[ 49 - k = 49 \implies k = 0 \] The radius of the sphere when \(k = 0\) is: \[ \sqrt{49 - 0} = 7 \] Thus, the radius of the sphere passing through the origin and concentric with the sphere S is 7.

📌 Hints / Properties Used:

Completing the square for equations of spheres.
Distance formula in three dimensions.

📊 Visual Diagram Suggestion:

Illustrate the sphere with center at (2, 3, 6) and radius of 7, along with a line connecting the center to the origin (0, 0, 0) depicting the radius.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Mathematics, Chapter 11 (Conic Sections).

Question 3 of 10 Authentic UPSC NDA Question

For the following two (02) items:

Suppose S is the sphere with the smallest radius that passes through the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1). On

which one of the following planes does the centre of S lie?

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Question 4 of 10 Authentic UPSC NDA Question

For the following two (02) items: Suppose S is the sphere with the smallest radius that passes through the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1).

What is the radius of S?

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Question 5 of 10 Authentic UPSC NDA Question

The centre of the sphere lies on the plane

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Question 6 of 10 Authentic UPSC NDA Question

Direction: Consider the following for the two (02) items
that follow:
Let 2x²+2y²+ 2z²+ 3x + 3y + 3z – 6 = 0 be a
sphere.

What is the diameter of the sphere?

Step-by-Step Logic & Solution

Question 7 of 10 Authentic UPSC NDA Question

If (1, –1, 2) and (2, 1, –1) are the end points of a diameter of a sphere x² + y² + z² + 2ux + 2vy + 2wz – 1 = 0, then what is u + v + w equal to?

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Question 8 of 10 Authentic UPSC NDA Question

What is the equation of the sphere concentric with the sphere x² + y² + z² − 2x − 6y − 8z − 5 = 0 and which passes through the origin?

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Question 9 of 10 Authentic UPSC NDA Question

For the next two (02) items:

Let A(1, −1, 2) and B(2, 1, −1) be the end points of the diameter of the sphere

x² + y² + z² + 2ux + 2vy + 2wz − 1 = 0.

If P(x, y, z) is any point on the sphere, then what is PA² + PB² equal to?

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Question 10 of 10 Authentic UPSC NDA Question

For the next two (02) items:

Let A(1, −1, 2) and B(2, 1, −1) be the end points of the diameter of the sphere

x² + y² + z² + 2ux + 2vy + 2wz − 1 = 0.

What is u + v + w equal to?

Step-by-Step Logic & Solution

So, from given equation of sphere we have

2a = –3, 2v = 0 and 2w = –1

⇒ a = –3/2, v = 0, w = –1/2

So, u + v + w = 4/2 = –2

Equation of a sphere Questions

Solution