Question

Find the path traced by the centre of the sphere which touches the lines x/y=y/-1z/2
and 2x=y,y-z=0 .

Answer 1

Answer on the Question #46095 – Math – Analytic Geometry

The path traced by the centre of the sphere is the locus. The radius of the sphere equals to the distance between the centre of the sphere and the points on following lines.

So, let $r_1 = \frac{x}{y} = \frac{y}{-1} = \frac{z}{2}$; $r_2 = \frac{x}{1} = \frac{y}{2} = \frac{z}{2}$;

Equation of the Sphere:

$$R^2 = (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 \quad (x_1, y_1, z_1 - \text{centre of the sphere})$$ $$R^2 = (x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2$$

So,

$$\begin{array}{l} (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = (x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2 \\ (x + r_1^2)^2 + (y + r_1)^2 + (z - 2r_1)^2 = (x - r_2)^2 + (y - 2r_2)^2 + (z - 2r_2)^2 \\ \end{array}$$

Open the brackets:

$$r_1^4 + 2r_1^2x + 2r_1y + r_1^2 - 4r_1z + 4r_1^2 = -2r_2x + r_2^2 - 4r_2y + 4r_2^2 - 4r_2z + 4r_2^2$$

So at the end we have:

$$(2r_1^2 + 2r_2)x + (2r_1 + 4r_2)y + (4r_2 - 4r_1)z + (r_1^4 + 5r_1^2 - 9r_2^2) = 0$$

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