Question

The radius of the sphere \[ x^2 + y^2 + z^2 = 49, \quad 2x + 3y - z - 5\sqrt{14} = 0 \] is:

• A. \( \sqrt{6} \)
• B. \( \sqrt{2} \)
• C. \( \sqrt{4/6} \)
• D. \( \sqrt{6} \)

Answer 1

The Correct Option is A

Step 1: Identify the radius of the given sphere
The equation of the sphere is: \[ x^2 + y^2 + z^2 = 49 \] So, the radius of the sphere is: \[ R = \sqrt{49} = 7 \]

Step 2: Find the distance of the plane from the center of the sphere
The center of the sphere is: \[ (0, 0, 0) \] The plane is: \[ 2x + 3y - z - 5\sqrt{14} = 0 \] Distance of a point \((x_1,y_1,z_1)\) from a plane \(Ax+By+Cz+D=0\) is: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] Substituting values: \[ d = \frac{| -5\sqrt{14} |}{\sqrt{2^2 + 3^2 + (-1)^2}} = \frac{5\sqrt{14}}{\sqrt{14}} = 5 \]

Step 3: Find the radius of the circle of intersection
The radius \(r\) of the circle formed by the intersection of a sphere and a plane is: \[ r = \sqrt{R^2 - d^2} \] Substitute values: \[ r = \sqrt{7^2 - 5^2} = \sqrt{49 - 25} = \sqrt{24} = \sqrt{6} \]

Final Answer:
The required radius is: \[ \boxed{\sqrt{6}} \]