Question

Find the centre and radius of the circle in which sphere x2 + y2 + z2 + 2x - 2y - 4z - 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0.
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Answer 1

Given info : sphere x² + y² + z² + 2x - 2y - 4z - 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0.

To find : the centre and the radius of the circle.

solution : sphere x² + y² + z² + 2x - 2y - 4z - 19 = 0

⇒x² + 2x + 1 + y² - 2y + 1 + z² - 4z + 4 = 1 + 1 + 4 + 19

⇒(x + 1)² + (y - 1)² + (z - 2)² = 25 = 5²

so, radius of sphere , AP = 5 , centre of sphere, B = (-1, 1, 2).

see figure,

sphere is cut by the plane x + 2y + 2z + 7 = 0 at P

distance from centre of sphere to plane , PB = |-1 + 2 × 1 + 2 × 2 + 7|/√(1² + 2² + 2²) = 4

from ∆APB,

AB² = AP² - PB²

⇒AB² = (5)² - (4)² = 3²

⇒AB = 3

therefore the centre and the radius of circle are (-1, 1, 2) and 3 units respectively.

Answer 2

Answer:

Answer is AB =5*2-3*2=3

so AB=3