Question

Find the equations of those tangent planes to the sphere x^2+y^2+Z^2+2x-4y+6z-7=0 which intersect in the line 6x-3y-23=0,3z+2=0

and

Under what conditions on α do the spheres x^2 + y^2+ z^2+x+2z+1=0 intersect each other at an angle of 45 degree ?

Answer 1

Answer on Question #82459 – Math – Analytic Geometry

Question

Find the equations of those tangent planes to the sphere

$$x^2 + y^2 + z^2 + 2x - 4y + 6z - 7 = 0$$

which intersect in the line $6x - 3y - 23 = 0, 3z + 2 = 0$.

Solution

The equation of the sphere

$$x^2 + y^2 + z^2 + 2x - 4y + 6z - 7 = 0$$ $$(x^2 + 2x + 1) + (y^2 - 4y + 4) + (z^2 + 6z + 9) - 1 - 4 - 9 - 7 = 0$$ $$(x + 1)^2 + (y - 2)^2 + (z + 3)^2 = 21$$

Its centre is $(-1, 2, -3)$ and radius $= \sqrt{21}$.

The equation of the lines

$$6x - 3y - 23 = 0, \quad 3z + 2 = 0$$

Any plane through this line is

$$(6x - 3y - 23) + k(3z + 2) = 0$$

or

$$6x - 3y + 3kz + 2k - 23 = 0$$

It will touch the sphere (1) if perpendicular from centre on plane (2) = radius of sphere (1)

$$\frac{|6(-1) - 3(2) + 3k(-3) + 2k - 23|}{\sqrt{(6)^2 + (-3)^2 + (3k)^2}} = \sqrt{21}$$ $$|7k + 35| = \sqrt{21} \sqrt{45 + 9k^2}$$ $$49(k + 5)^2 = 21(45 + 9k^2)$$ $$7k^2 + 70k + 175 = 135 + 27k^2$$ $$20k^2 - 70k - 40 = 0$$ $$2k^2 - 7k - 4 = 0$$ $$k = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(-4)}}{2(2)} = \frac{7 \pm 9}{4}$$ $$k_1 = \frac{7 - 9}{4} = -\frac{1}{2}$$ $$k_2 = \frac{7 + 9}{4} = 4$$

Putting these values of $k$ in (2), we get

$$6x - 3y + 3\left(-\frac{1}{2}\right)z + 2\left(-\frac{1}{2}\right) - 23 = 0$$ $$6x - 3y - \frac{3}{2}z - 24 = 0$$ $$4x - 2y - z - 16 = 0$$ $$6x - 3y + 3(4)z + 2(4) - 23 = 0$$ $$2x - y + 4z - 5 = 0$$

Answer: $4x - 2y - z - 16 = 0$ and $2x - y + 4z - 5 = 0$.

Question

Under what conditions on $\alpha$ do the spheres $x^{2} + y^{2} + z^{2} + \alpha x - y = 0$ and $x^{2} + y^{2} + z^{2} + x + 2z + 1 = 0$

intersect each other at an angle of 45 degree?

Solution

$$S_{1}: x^{2} + y^{2} + z^{2} + \alpha x - y = 0$$

or

$$S_{1} \colon \left(x + \frac{\alpha}{2}\right)^{2} + \left(y - \frac{1}{2}\right)^{2} + z^{2} = \frac{\alpha^{2} + 1}{4}$$

Its centre is $\left(-\frac{\alpha}{2}, \frac{1}{2}, 0\right)$ and radius $= \frac{\sqrt{\alpha^{2} + 1}}{2}$.

$$S_{2}: x^{2} + y^{2} + z^{2} + x + 2z + 1 = 0$$

or

$$S_{2} \colon \left(x + \frac{1}{2}\right)^{2} + y^{2} + (z + 1)^{2} = \frac{1}{4}$$

Its centre is $\left(-\frac{1}{2}, 0, -1\right)$ and radius $= \frac{1}{2}$.

The intersection plane is obtained as $S_{2} - S_{1} = 0$

$$x^{2} + y^{2} + z^{2} + x + 2z + 1 - (x^{2} + y^{2} + z^{2} + \alpha x - y) = 0$$

or

$$(1 - \alpha)x + y + 2z + 1 = 0$$

The sphere's center distance

$$d = \sqrt{\left(-\frac{\alpha}{2} - \left(-\frac{1}{2}\right)\right)^{2} + \left(\frac{1}{2} - 0\right)^{2} + (0 - (-1))^{2}}$$

Consider the intersection point as the first vertex and the sphere centers as the remaining two triangle vertices. The Law of Cosines

$$d^{2} = r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2}\cos\theta$$

If $\theta = 45{}^{\circ}$

$$\frac{(1 - \alpha)^{2}}{4} + \frac{1}{4} + 1 = \frac{\alpha^{2} + 1}{4} + \frac{1}{4} - 2\left(\frac{\sqrt{\alpha^{2} + 1}}{2}\right)\left(\frac{1}{2}\right)\cos 45{}^{\circ}$$ $$\sqrt{2}\sqrt{\alpha^{2} + 1} = \alpha^{2} + 1 - 1 + 2\alpha - \alpha^{2} - 4$$ $$\sqrt{2}\sqrt{\alpha^{2} + 1} = 2\alpha - 4, 2\alpha - 4 > 0 \Rightarrow \alpha > 2$$ $$2(\alpha^{2} + 1) = 4(\alpha - 2)^{2}$$ $$2\alpha^{2} - 8\alpha + 8 - \alpha^{2} - 1 = 0$$ $$\alpha^{2} - 8\alpha + 7 = 0$$ $$\begin{array}{l} \alpha = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(7)}}{2(1)} = 4 \pm 3 \\ \alpha_1 = 4 - 3 = 1, \alpha_2 = 4 + 3 = 7 \\ \text{Since } \alpha > 2, \text{ we take } \alpha = 7. \end{array}$$

Answer: $\alpha = 7$.

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