find the equation of the sphere which pass through the circle x^2+y^2+z^2=5,x+2y+3z=3, and touch the plane 4x+3y=15
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Answer:
Equation to the sphere passes through the circle x
2
+y
2
+z
2
=5,x+2y+3z=3 is given by,
x
2
+y
2
+z
2
−5+λ(x+2y+3z−3)=0
⇒x
2
+y
2
+z
2
+λx+2λy+3λz−(3λ+5)=0
Given plane 4x+3y−15=0 touches this sphere,
∣
∣
∣
∣
∣
∣
5
4(
2
−λ
)+3(−λ)−15
∣
∣
∣
∣
∣
∣
=
(
2
λ
)
2
+λ
2
+(
2
3λ
+3λ+5)
2
⇒5λ
2
−6λ−8=0⇒λ=2,−4/5
Hence, the sphere is x
2
+y
2
+z
2
+2x+4y+6z−11=0
There will be one more sphere corresponding to another value of λ.
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