The plane `x/a+y/b+z/c=1` meets the coordinate axes at A,B and C respectively. Find the equation of the sphere OABC.

Options

A. `x^(2)+y^(2)+z^(2)+az+by+cz=0`
B. `x^(2)+y^(2)+z^(2)+2az+2by+2cz=0`
C. `x^(2)+y^(2)+z^(2)-ax-by-cz=0`
D. None of these

Answer

To find the equation of the sphere OABC, where O is the origin and A, B, and C are the points where the plane \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \) meets the coordinate axes, we will follow these steps:

### Step 1: Find the coordinates of points A, B, and C
The plane intersects the coordinate axes at points A, B, and C.

- To find point A (where the plane intersects the x-axis), set \( y = 0 \) and \( z = 0 \):
\[ \frac{x}{a} + 0 + 0 = 1 \implies x = a \implies A(a, 0, 0) \]

- To find point B (where the plane intersects the y-axis), set \( x = 0 \) and \( z = 0 \):
\[ 0 + \frac{y}{b} + 0 = 1 \implies y = b \implies B(0, b, 0) \]

- To find point C (where the plane intersects the z-axis), set \( x = 0 \) and \( y = 0 \):
\[ 0 + 0 + \frac{z}{c} = 1 \implies z = c \implies C(0, 0, c) \]

### Step 2: Find the center of the sphere
The center of the sphere OABC is the midpoint of the line segment joining the origin O(0, 0, 0) and the centroid of triangle ABC. The coordinates of the centroid G of triangle ABC can be calculated as:
\[ G\left(\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \]
Thus, the center of the sphere is at the midpoint between O and G:
\[ \text{Center} = \left(\frac{0 + \frac{a}{3}}{2}, \frac{0 + \frac{b}{3}}{2}, \frac{0 + \frac{c}{3}}{2}\right) = \left(\frac{a}{6}, \frac{b}{6}, \frac{c}{6}\right) \]

### Step 3: Calculate the radius of the sphere
The radius of the sphere is the distance from the center to any of the points A, B, or C. We can calculate the distance from the center to point A:
\[ \text{Radius} = \sqrt{\left(a - \frac{a}{6}\right)^2 + \left(0 - \frac{b}{6}\right)^2 + \left(0 - \frac{c}{6}\right)^2} \]
Calculating this gives:
\[ = \sqrt{\left(\frac{5a}{6}\right)^2 + \left(-\frac{b}{6}\right)^2 + \left(-\frac{c}{6}\right)^2} = \sqrt{\frac{25a^2}{36} + \frac{b^2}{36} + \frac{c^2}{36}} = \sqrt{\frac{25a^2 + b^2 + c^2}{36}} = \frac{\sqrt{25a^2 + b^2 + c^2}}{6} \]

### Step 4: Write the equation of the sphere
The equation of a sphere with center \( (h, k, l) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \]
Substituting \( h = \frac{a}{6} \), \( k = \frac{b}{6} \), \( l = \frac{c}{6} \), and \( r = \frac{\sqrt{25a^2 + b^2 + c^2}}{6} \):
\[ \left(x - \frac{a}{6}\right)^2 + \left(y - \frac{b}{6}\right)^2 + \left(z - \frac{c}{6}\right)^2 = \left(\frac{\sqrt{25a^2 + b^2 + c^2}}{6}\right)^2 \]
This simplifies to:
\[ \left(x - \frac{a}{6}\right)^2 + \left(y - \frac{b}{6}\right)^2 + \left(z - \frac{c}{6}\right)^2 = \frac{25a^2 + b^2 + c^2}{36} \]

### Final Equation of the Sphere
Thus, the equation of the sphere OABC is:
\[ \left(x - \frac{a}{6}\right)^2 + \left(y - \frac{b}{6}\right)^2 + \left(z - \frac{c}{6}\right)^2 = \frac{25a^2 + b^2 + c^2}{36} \]