Question #132947
Find the equations of the spheres which pass through the circle x²+y²+z²=0, x+2y+3z=0 and touch the plane 4x+3y=15
1
Expert's answer
2020-09-15T11:01:52-0400

The equation of a sphere is:

x2+y2+z2a2λ(x+2y+3z)=0x²+y²+z²-a-2\lambda(x+2y+3z)=0

(x22λx+λ2)λ2+(y24λy+4λ2)4λ2+(z26λz+9λ2)9λ2=a(x^2-2\lambda x+\lambda^2)-\lambda^2+(y^2-4\lambda y+4\lambda^2)-4\lambda^2+(z^2-6\lambda z+9\lambda^2)-9\lambda^2=a

(xλ)2+(y2λ)2+(z3λ)2=a+14λ2(x-\lambda)^2+(y-2\lambda)^2+(z-3\lambda)^2=a+14\lambda^2

The radius of the sphere:

R=a+14λ2R=\sqrt{a+14\lambda^2}

The center is:

(λ,2λ,3λ)(\lambda,2\lambda,3\lambda)

The distance from center to the tangent plane:

d=4λ+6λ1542+32=2λ3d=\frac{|4\lambda+6\lambda-15|}{\sqrt{4^2+3^2}}=|2\lambda-3|

We have:

R=dR=d

Then:

a+14λ2=2λ3\sqrt{a+14\lambda^2}=|2\lambda-3|

a+14λ2=4λ212λ+9a+14\lambda^2=4\lambda^2-12\lambda+9

10λ2+12λ+a9=010\lambda^2+12\lambda+a-9=0

λ=12±14440(a9)20=6±3810(a9)10\lambda=\frac{-12\pm\sqrt{144-40(a-9)}}{20}=\frac{-6\pm\sqrt{38-10(a-9)}}{10}

Substitute λ\lambda in the equation of the sphere, and we get equations of two different spheres.



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