Solution.
Find the equation of normal to the plane:
4x−8=3y−5=0z−4=k, where k is some number.
Then
C(4k+8;3k+5;4)− center of required spheres.
Radius of required spheres:
R=(4k+8−8)2+(3k+5−5)2+(4−4)2==25k2=5k. Sphere x2+y2+z2=1 has center (0;0;0) and radius 1.
We will have equation:
(1+5k)2=(5k)2+105, where 105is distance between (0;0;0) and (8;5;4).
From here
1+10k+25k2=25k2+105,k=10.4 So, (49.6;36.2;4) and (−33.6;−26.2;4) are the centres of spheres.
Answer. (−33.6;−26.2;4),(49.6;36.2;4).
Comments
Dear Naushad ahmad, we received answers provided in a solution of the question.
But sir answer is (4, 2 , 4) and (64/21, 27/21, 4) . Please comment
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