The equation of the sphere one of whose diameter has end points (1,2,4) and (3,0,2) is
A
x2+y2+z2+4x+6y+8z+11=0
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B
x2+y2+z2+4x+4y+8z−11=0
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C
x2+y2+z2−4x−2y−6z+11=0
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D
x2+y2+z2−4x−2y+6z−11=0
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Solution
The correct option is Cx2+y2+z2−4x−2y−6z+11=0 End point of diameter of sphere are (1,2,4) and (3,0,2). So centre of sphere is (2,1,3) and radius is =12√(1−3)2+(2−0)2+(4−2)2=√3 Hence, equation of required sphere is (x−2)2+(y−1)2+(z−3)2=3 ⇒x2+y2+z2−4x−2y−6y+11=0