The radius of the circle in which the sphere | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2w

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The equation of the sphere is $x^2 + y^2 + z^2 + 2x - 2y - 4z - 19 = 0$.Comparing this with the general equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$,we get the center $C = (-u, -v, -w) = (-1, 1, 2)$ and the radius $R = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1 + 1 + 4 + 19} = \sqrt{25} = 5$.The perpendicular distance $p$ from the center $C(-1, 1, 2)$ to the plane $x + 2y + 2z + 7 = 0$ is given by $p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1 + 4 + 4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^2 - p^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.

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

Similar Questions

The shortest distance from the origin to a variable point on the sphere $(x-2)^2+(y-3)^2+(z-6)^2=1$ is

$A$ plane passes through a fixed point $(p, q, r)$ and cuts the axes at $A, B, C$. Then the locus of the centre of the sphere $OABC$ is

The locus of a point $P(x, y, z)$ at which the line segment joining the points $A(-3, 1, 2)$ and $B(1, -2, 4)$ subtends a right angle is:

What is the center of the sphere passing through the four points $(0, 0, 0), (0, 2, 0), (1, 0, 0),$ and $(0, 0, 4)$?

If $(2, 3, 5)$ is one end of a diameter of the sphere $x^2 + y^2 + z^2 - 6x - 12y - 2z + 20 = 0$,then the coordinates of the other end of the diameter are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module