Answer to Question #166776 in Analytic Geometry for Anshika

Question #166776

Find the centres of the two spheres which touch the plane 4x+3y=47 at the point (8,5,4) and the sphere x^2+y^2+z^2=1


1
Expert's answer
2021-03-04T15:23:06-0500

Solution.

Find the equation of normal to the plane:


x84=y53=z40=k,\frac{x-8}{4}=\frac{y-5}{3}=\frac{z-4}{0}=k,

where kk is some number.

Then

C(4k+8;3k+5;4)C(4k+8;3k+5;4)-

center of required spheres.

Radius of required spheres:


R=(4k+88)2+(3k+55)2+(44)2==25k2=5k.R=\sqrt{(4k+8-8)^2+(3k+5-5)^2+(4-4)^2}=\newline =\sqrt{25k^2}=5k.

Sphere x2+y2+z2=1x^2+y^2+z^2=1 has center (0;0;0)(0;0;0) and radius 1.1.

We will have equation:

(1+5k)2=(5k)2+105, where 105(1+5k)^2=(5k)^2+105, \newline \text{ where } 105is distance between (0;0;0) and (8;5;4).

From here

1+10k+25k2=25k2+105,k=10.41+10k+25k^2=25k^2+105, \newline k=10.4

So, (49.6;36.2;4)(49.6;36.2;4) and (33.6;26.2;4)(-33.6;-26.2;4) are the centres of spheres.

Answer. (33.6;26.2;4),(49.6;36.2;4).(-33.6;-26.2;4), (49.6;36.2;4).



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
09.03.21, 14:15

Dear Naushad ahmad, we received answers provided in a solution of the question.

Naushad ahmad
09.03.21, 08:34

But sir answer is (4, 2 , 4) and (64/21, 27/21, 4) . Please comment

Leave a comment