The equation of the sphere touching the three | vedclass

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

Question

(B) sphere touching the three coordinate planes has its center at $(a, a, a)$ and radius $a$ (assuming the sphere is in the first octant).The standard

Vedclass · Accepted Answer

$x^2 + y^2 + z^2 - 2a(x + y + z) + 2a^2 = 0$

Vedclass · Answer

(B) sphere touching the three coordinate planes has its center at $(a, a, a)$ and radius $a$ (assuming the sphere is in the first octant).The standard equation of a sphere with center $(h, k, l)$ and radius $r$ is $(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$.Substituting the center $(a, a, a)$ and radius $r = a$,we get:$(x - a)^2 + (y - a)^2 + (z - a)^2 = a^2$Expanding the squares:$(x^2 - 2ax + a^2) + (y^2 - 2ay + a^2) + (z^2 - 2az + a^2) = a^2$Simplifying the equation:$x^2 + y^2 + z^2 - 2ax - 2ay - 2az + 3a^2 = a^2$$x^2 + y^2 + z^2 - 2a(x + y + z) + 2a^2 = 0$Thus,the required equation is $x^2 + y^2 + z^2 - 2a(x + y + z) + 2a^2 = 0$.

The equation of the sphere touching the three coordinate planes is

Similar Questions

If $A(3, -2, 2)$ and $B(2, \lambda+1, 5)$ are the end points of the diameter of a circle and the point $P(5, 6, -1)$ lies on the circle,then $\lambda=$

The coordinate of a point equidistant from the points $(0, 0, 0), (a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ is

The plane $x + 2y - z = 4$ cuts the sphere $x^2 + y^2 + z^2 - x + z - 2 = 0$ in a circle of radius:

Find the radius of the circle formed by the intersection of the sphere $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$ and the plane $x + 2y + 2z = 15$.

If two spheres of radii $r_1$ and $r_2$ cut orthogonally,then the radius of the common circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module