Question

Find the equations of the spheres which pass through the circle x² +y² +z² = 9,2x+2y−7 = 0 and the touch the plane x−y +z +3 = 0

Answer 1

1. Suppose that the sphere O₁ passes through the circle O₂ (x² + y² + z² = 9, 2x + 2y - 7 = 0).

Then its equation is:

x² + y² + z² - k(2x + 2y - 7) = 9;

(x - k)² + (y - k)² + z² = 2k² - 7k + 9,

where k is any real number.

Its center is O₁(k, k, 0) and the radius is R₁ = $\sqrt{2k^2 - 7k + 9}$

2. Since the sphere touches the plane b (x - y + z + 3 = 0), then the distance O₁H 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 = 1, b = -1, c = 1, d = 3, x₀ = k, y₀ = k, z₀ = 0, then:

$\frac{d = |1 * k - 1 * k + 1 * 0 + 3|/}{\sqrt{1^2 + (-1)^2 + 1^2}}$ = $3 \over \sqrt3$ = $\sqrt3$ .

4. From R₁ = d:

$\sqrt{2k^2 - 7k + 9}=\sqrt3$;

2k² - 7k + 9 = 3;

2k² - 7k + 6 = 0;

D = 7² - 4 * 2 * 6 = 49 - 48 = 1;

$\frac{k = (7 ± 1)}{4};$

k₁ = 1.5; k₂ = 2.

The equations of two spheres are:

(x - k)² + (y - k)² + z² = 2k² - 7k + 9,

1) k = 1.5; (x - 1.5)² + (y - 1.5)² + z² = 3;

2) k = 2; (x - 2)² + (y - 2)² + z² = 3.

Answer: (x - 1.5)² + (y - 1.5)² + z² = 3; (x - 2)² + (y - 2)² + z² = 3.