Answer:
[tex]\mathbf{ x^2 + (y-8)^2 +(z)^2 = 32}[/tex]
Step-by-step explanation:
Given that:
The center of the sphere = (0,8,0)
[tex]The \ equation \ of \ the \ tangent \ plane \ is:[/tex]
= x - y = 0
x = y
However, the radius of the sphere is equal to the perpendicular distance of the plane from the center.
As such, the radius [tex]r = \Big | \dfrac{ax_1 +by_1+cz_1}{\sqrt{a^2 + b^2 +c^2}} \Big |[/tex]
[tex]r = \Big | \dfrac{1*0 +(-1)*8+0*0}{\sqrt{1^2 + (-1)^2 +0^2}} \Big |[/tex]
[tex]r = \Big | \dfrac{-8}{\sqrt{2}} \Big |[/tex]
[tex]r = \dfrac{8}{\sqrt{2}}[/tex]
[tex]r = \dfrac{4 \times 2}{\sqrt{2}}[/tex]
[tex]r = 4\sqrt{2}[/tex] units
Thus, the equation of the sphere [tex]r^2 = (x -x_1)^2 +(y-y_1)^2 +(z -z_1)^2[/tex]
[tex]\mathbf{\implies x^2 + (y-8)^2 +(z)^2 = 32}[/tex]
