Solution:
Given: Sphere passes through (0,0,0),(2,1,−1),(1,5,−4) and (−2,4,−6)
General equation of sphere, x2+y2+z2+2ux+2vy +2wz+d=0 .......(1)
Put (0,0,0) in the equation (1) we get,
0+0+0+0+0+0+d=0
⇒d=0
Now, equation becomes, x2+y2+z2+2ux+2vy +2wz=0 ......(2)
Put (2,1,−1) in equation ( 2 ) we get,
22+12+(−1)2+2u(2)+2v(1) +2w(−1)=0
⇒6+4u+2v−2w=0 .......(3)
Put (1, 5, -4) in equation (2)
12+52+(−4)2+2u(1)+2v(5) +2w(−4)=0
⇒42+2u+10v−8w=0 .......(4)
Put (-2, 4, -6) in equation (2)
(−2)2+42+(−6)2+2u(−2)+2v(4) +2w(−6)=0
⇒ 56 - 4u + 8v - 12w = 0 ......(5)
Now, adding equation (3) and (5)
62 + 10v - 14w = 0 .......(6)
Divide equation (5) and add it to equation (4)
28 - 2u + 4v - 6w + 42 + 2u + 10v - 8w = 0
⇒ 70 + 14v - 14w = 0 .......(7)
Now, subtracting (6) from (7)
8 + 4v = 0
⇒ v = -2
Put v = -2 in equation (7)
70 + 14(-2) - 14w = 0
⇒ 14w = 42
⇒ w = 3
Put v = -2 and w = 3 in equation (3)
6 + 4u + 2(-2) - 2(3) = 0
⇒ 4u = 4
⇒ u = 1
Equation of circle is given as,
x2+y2+z2+2x−4y+6z=0
Radius of circle, r=√u2+v2+z2−d
⇒r=√12+(−2)2+32
⇒r=√14
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