Q.384. Center and radius of the circle in 3-D.

Question 384.
Find the coordinates of center and radius of the circle of intersection of 
the sphere x²+y²+z²=9 with the plane 3x+4y+5z=5,

Answer 384.
The equation of the family of spheres through the given sphere and the given plane is
x^2 + y^2 + z^2 - 9 + k (3x + 4y + 5k - 5) = 0.

The centre of the intersecting circle is the centre of the sphere from the above family which lies on the plane.
The centre of the sphere = (-3k/2, - 4k/2, - 5k/2)

Plugging in the eqn. of the plane
=> - 9k/2 - 16k/s - 25k/2 - 5 = 0
=> k = -1/5
=> eqn. of the sphere containing the intersecting circle as its large circle is
x^2 + y^2 + z^2 - 9 - (1/5) (3x + 4y + 5z - 5) = 0
Its centre is (3/10, 2/5, 1/2)
Its radius
= √[(3/10)^2 + (4/10)^2 + (5/10)^2 + 9 - 1]
= √[1/2 + 8]
= √(17/2)

Alternate method to find the radius is
radius of the sphere
r = 3
perpendicular distance from the centre (0, 0, 0) of the sphere to the plane
p = 5/√(3^2 + 4^2 + 5^2) = 1/√2
=> radius of the circle of intersection
= √(r^2 - p^2)
= √(9 - 1/2)
= √(17/2).

Link to YA!