Question

If a plane passes through a fixed point \((a,b,c)\) and cuts the axes in \(A,B,C\), then the locus of the centre of the sphere \(OABC\) is

• A. \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=1\)
• B. \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\)
• C. \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
• D. \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=2\)

Hint:

For sphere through origin and intercepts \((\alpha,0,0),(0,\beta,0),(0,0,\gamma)\), centre is \((\alpha/2,\beta/2,\gamma/2)\).

Answer 1

The Correct Option is B

Concept:
Let the plane cut the coordinate axes at \[ A(\alpha,0,0),\quad B(0,\beta,0),\quad C(0,0,\gamma) \] Then its intercept form is \[ \frac{X}{\alpha}+\frac{Y}{\beta}+\frac{Z}{\gamma}=1 \]

Step 1: Since the plane passes through \((a,b,c)\). Substitute \[ X=a,\quad Y=b,\quad Z=c \] in the plane equation: \[ \frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}=1 \]

Step 2: Equation of sphere through \(O,A,B,C\). The sphere passing through origin and intercept points has equation \[ X^2+Y^2+Z^2-\alpha X-\beta Y-\gamma Z=0 \] Its centre is \[ \left(\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right) \] Let the centre be \[ (x,y,z) \] Then, \[ x=\frac{\alpha}{2},\quad y=\frac{\beta}{2},\quad z=\frac{\gamma}{2} \] So, \[ \alpha=2x,\quad \beta=2y,\quad \gamma=2z \]

Step 3: Substitute in the condition. \[ \frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}=1 \] Using \[ \alpha=2x,\quad \beta=2y,\quad \gamma=2z \] we get \[ \frac{a}{2x}+\frac{b}{2y}+\frac{c}{2z}=1 \] Multiply by \(2\): \[ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2 \]

Step 4: Final answer. \[ \boxed{\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2} \]