wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question


Equation to the sphere which passes through the circle x2+y2+z2=5,x+2y+3z=3 and touch the plane 4x+3y− 15=0

A
x2+y2+z2+2x+4y+6z+11=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x2+y2+z22x4y+6z11=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2+y2+z22x4y6z+11=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2+y2+z2+2x+4y+6z11=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is B x2+y2+z2+2x+4y+6z11=0
Equation to the sphere passes through the circle x2+y2+z2=5,x+2y+3z=3 is given by,
x2+y2+z25+λ(x+2y+3z3)=0
x2+y2+z2+λx+2λy+3λz(3λ+5)=0
Given plane 4x+3y15=0 touches this sphere,
∣ ∣ ∣4(λ2)+3(λ)155∣ ∣ ∣=(λ2)2+λ2+(3λ2+3λ+5)2
5λ26λ8=0λ=2,4/5
Hence, the sphere is x2+y2+z2+2x+4y+6z11=0
There will be one more sphere corresponding to another value of λ.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon