Answer to Question #132947 in Analytic Geometry for Harshita

Question

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

Answer 1

The equation of a sphere is:

$x²+y²+z²-a-2\lambda(x+2y+3z)=0$

$(x^2-2\lambda x+\lambda^2)-\lambda^2+(y^2-4\lambda y+4\lambda^2)-4\lambda^2+(z^2-6\lambda z+9\lambda^2)-9\lambda^2=a$

$(x-\lambda)^2+(y-2\lambda)^2+(z-3\lambda)^2=a+14\lambda^2$

The radius of the sphere:

$R=\sqrt{a+14\lambda^2}$

The center is:

$(\lambda,2\lambda,3\lambda)$

The distance from center to the tangent plane:

$d=\frac{|4\lambda+6\lambda-15|}{\sqrt{4^2+3^2}}=|2\lambda-3|$

We have:

$R=d$

Then:

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

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

$10\lambda^2+12\lambda+a-9=0$

$\lambda=\frac{-12\pm\sqrt{144-40(a-9)}}{20}=\frac{-6\pm\sqrt{38-10(a-9)}}{10}$

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