Question

find the centres of two spheres which touch
the plane 4x+ 3y = 47 at the point (8,5,4) and
the sphere x² + y² +z²=1.​

Answer 1

(49.6, 36.2, 4) and (−33.6, −26.2, 4) are the centers of the two spheres.

Given:

The centers of two spheres touch the plane 4x+ 3y = 47 at the point (8,5,4) and the sphere x² + y² +z²=1.

To Find:

The centers of the two spheres.

Solution:

The equation of normal to the plane is given by:

$\frac{x -x_{0} }{a} = \frac{y-y_{o} }{b} = \frac{z-z_{o} }{c} = k$, where k is some constant.

The equation of a plane is given by ax + by + cz = d

The given plane is 4x+ 3y = 47.

By comparing we get (a, b, c) = (4, 3, 0)

The given point is ($x_{0} , y_{0} , z_{0}$) = (8, 5, 4)

Hence the equation of normal to the plane becomes,

$\frac{x -8 }{4} = \frac{y-5 }{3} = \frac{z-4 }{0} = k$

⇒ x = 4k+8, y = 3k+5, z = 4.

Now, the center of the spheres is C (4k+8, 3k+5, 4)

Now, the radius of required spheres R = $\sqrt{(x-x_{0} )^{2} + (y-y_{0} )^{2} + (z-z_{0} )^{2} }$

⇒ R = $\sqrt{(4k + 8 - 8)^{2} + (3k + 5 - 5)^{2} + (4 -4)^{2}}$ = $\sqrt{25k^{2}}$ = ±5k = 5(±k)

Now, the sphere x² + y² +z² = 1 will have center (0, 0, 0) and radius 1.

The distance between (0, 0, 0) and (8, 5, 4) is $\sqrt{8^{2} + 5^{2} + 4^{2} }$ = 105

We will have equation:

(1 + 5(±k))² = (5(±k))² + 105

⇒ 1 ± 10k + 25k² = 25k² + 105

⇒ k = ±10.4

Substituting this in C (4k+8, 3k+5, 4), we get  (49.6, 36.2, 4) and (−33.6, −26.2, 4) as the centers of the spheres.

∴ (49.6, 36.2, 4) and (−33.6, −26.2, 4) are the centers of the two spheres.

#SPJ1