Answer to Question #146839 in Analytic Geometry for Dhruv rawat

Question

Find the equations of the spheres which pass through the circle x^2+y^2+z^2-6x+z+6=0, x-y=0 and touch the plane z=0

Answer 1

Suppose that the sphere $O_1$ passes though the circle $O_2$ (x²+y²+z²–6x+z+6=0, x–y=0)

Then its equation is:

$x^2 + y^2 + z^2 -6x+z+6-k(x-y) = 0$

where k is any real number.

$x^2-x(k+6)+y^2+ky+z^2+z+6=0$

$x^2-2x(\frac{k}{2}+3)+(\frac{k}{2}+3)^2+y^2+2y\frac{k}{2}+(\frac{k}{2})^2+z^2+2z\frac{1}{2}+(\frac{1}{2})^2=(\frac{k}{2}+3)^2+(\frac{k}{2})^2+(\frac{1}{2})^2$

$(x-(\frac{k}{2}+3))^2+(y+\frac{k}{2})^2+(z+\frac{1}{2})^2=\frac{k^2}{2}+3k+3\frac{1}{4}$

Its center is $O_1(\frac{k}{2}+3, -\frac{k}{2}, -\frac{1}{2})$ and the radius is $R_1=\sqrt{\frac{k^2}{2}+3k+3\frac{1}{4}}$

Since the sphere touches the plane b (z = 0), then the distance from the point O₁ to the plane is equal to the radius R₁. Point-plane distance is determined by the formula:

Plane P: ax + by + cz + d = 0;

Point O: (x₀, y₀, z₀);

$$d=\frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

If a = 0, b = 0, c = 1, d = 0, x₀ = k/2 + 3, y₀ = -k/2, z₀ = -1/2, then:

$d=\frac{|-\frac{1}{2}\cdot1|}{\sqrt1}=\frac{1}{2}$

from R₁ = d:

$\sqrt{\frac{k^2}{2}+3k+3\frac{1}{4}}=1$

$k^2+6k+4\frac{1}{2}=0$

$k=-3\pm\frac{3\sqrt2}{2}$

The equations of two spheres are:

$(x-(\frac{k}{2}+3))^2+(y+\frac{k}{2})^2+(z+\frac{1}{2})^2=\frac{k^2}{2}+3k+3\frac{1}{4}$

1) $k=-3+\frac{3\sqrt2}{2}$ :

$(x-(\frac{-3+\frac{3\sqrt2}{2}}{2}+3))^2+(y+\frac{-3+\frac{3\sqrt2}{2}}{2})^2+(z+\frac{1}{2})^2=1$

$(x-\frac{3\sqrt2+6}{4})^2+(y+\frac{3\sqrt2-6}{4})^2+(z+\frac{1}{2})^2=1$

2) $k=-3-\frac{3\sqrt2}{2}$ :

$(x-(\frac{-3-\frac{3\sqrt2}{2}}{2}+3))^2+(y+\frac{-3-\frac{3\sqrt2}{2}}{2})^2+(z+\frac{1}{2})^2=1$

$(x-\frac{6-3\sqrt2}{4})^2+(y-\frac{6+3\sqrt2}{4})^2+(z+\frac{1}{2})^2=1$

Answer:

$(x-\frac{3\sqrt2+6}{4})^2+(y+\frac{3\sqrt2-6}{4})^2+(z+\frac{1}{2})^2=1$

$(x-\frac{6-3\sqrt2}{4})^2+(y-\frac{6+3\sqrt2}{4})^2+(z+\frac{1}{2})^2=1$