Q.221. To find equation of a sphere under given conditions.

Question 221.
Find the equation of the sphere which touches the plane 3x+2y-z+2=0 at the point ( 1, -2, 1) and also cuts orthogonally the sphere x^2 + y^2 + z^2 - 4x + 6y + 4 = 0.

Answer 221.
Let C (a, b, c) be the center of the required sphere and r = its radius.

Direction CP, where P is the point of contact (1, -2, 1) is along the normal (3, 2, -1) of the tangent plane 3x + 2y - z + 2 = 0
=> (a - 1) / 3 = (b + 2) / 2 = (c - 1) / (-1) = k
=> a = 3k +1, b = 2k - 2 and c = 1 - k

For the given sphere, x^2 + y^2 + z^2 - 4x + 6y + 4 = 0,
center is C' = (2, - 3, 0) and radius = 3

As the required sphere cuts the given sphere orthogonally,
(distance between their centers)^2 = sum of the squares of their radii
=> (a - 2)^2 + (b + 3)^2 + c^2 = 9 + r^2 ... (1)

r^2 = CP^2 = (a - 1)^2 + (b + 2)^2 + (c - 1)^2

Plugging this value of r^2 in eqn. (1),
(a - 2)^2 + (b + 3)^2 + c^2 = 9 + (a - 1)^2 + (b + 2)^2 + (c - 1)^2
=> a - b - c + 1 = 0 ... (after simplification)

Plugging the values of a, b and c in terms of k,
=> 3k + 1 - 2k + 2 - 1 + k + 1 = 0
=> k = - 3/2

=> a = 3k + 1 = -9/2 + 1 = - 7/2,
b = 2k - 2 = - 3 - 2 = - 5 and
c = 1 - k = 1 + 3/2 = 5/2

Also, r^2
= (a - 1)^2 + (b + 2)^2 + (c - 1)^2
= (- 7/2 - 1)^2 + (- 5 + 2)^2 + (5/2 - 1)^2
= 126/4

=> the equation of the required sphere is
(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2
=> (x + 7/2)^2 + (y + 5)^2 + (z - 5/2)^2 = 126/4
=> x^2 + y^2 + z^2 + 7x + 10y - 5z + (49/4 + 100/4 + 25/4 - 126/4) = 0
=> x^2 + y^2 + z^2 + 7x + 10y - 5z + 12 = 0.
 
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