The radius of the circle in which the sphere x2+y2+z2+2x-2y-4z-19=0 is cut by the plane x + 2y + 2z + 7 = 0 is

Question

The radius of the circle in which the sphere x2+y2+z2+2x-2y-4z-19=0 is cut by the plane x + 2y + 2z + 7 = 0 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ec65b50ec3efde30337a28536f5a571b-.png" /> Centre of sphere = (-1,1,2) Radius of sphere =√1+1+4+19=5 Perpendicular distance from centre to the plane OC=d=|(-1+2+4+7/√1+4+4)|=(12/3)=4 AC2=AO2-OC2=52-42=9 ⇒ AC=3

Q. The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Centre of sphere $= (- 1, 1, 2)$
Radius of sphere $= 1 + 1 + 4 + 19 = 5$
Perpendicular distance from centre to the plane
$OC = d = ∣ ∣ \frac{- 1 + 2 + 4 + 7}{1 + 4 + 4} ∣ ∣ = \frac{12}{3} = 4$
$A C^{2} = A O^{2} - O C^{2} = 5^{2} - 4^{2} = 9$
$\Rightarrow A C = 3$

Q. The radius of the circle in which the sphere x2+y2+z2+2x−2y−4z−19=0 is cut by the plane x+2y+2z+7=0 is

Centre of sphere =(−1,1,2) Radius of sphere =1+1+4+19​=5 Perpendicular distance from centre to the plane OC=d=∣∣​1+4+4​−1+2+4+7​∣∣​=312​=4 AC2=AO2−OC2=52−42=9 ⇒AC=3

Q. The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Centre of sphere $= (- 1, 1, 2)$
Radius of sphere $= 1 + 1 + 4 + 19 = 5$
Perpendicular distance from centre to the plane
$OC = d = ∣ ∣ \frac{- 1 + 2 + 4 + 7}{1 + 4 + 4} ∣ ∣ = \frac{12}{3} = 4$
$A C^{2} = A O^{2} - O C^{2} = 5^{2} - 4^{2} = 9$
$\Rightarrow A C = 3$