Question:medium

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

Show Hint

For sphere through origin and intercepts \((\alpha,0,0),(0,\beta,0),(0,0,\gamma)\), centre is \((\alpha/2,\beta/2,\gamma/2)\).
  • \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=1\)
  • \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\)
  • \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
  • \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=2\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We are looking for the locus (path) of the center of a sphere that is defined by four points: the origin O and the intercepts A, B, C of a variable plane. This variable plane has the constraint that it always passes through a fixed point (a,b,c).

Step 2: Key Formula or Approach:

1. Let the equation of the plane be in intercept form: \(\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1\). The intercepts are \(A=(p,0,0)\), \(B=(0,q,0)\), \(C=(0,0,r)\). 2. The sphere passes through O(0,0,0), A, B, and C. The equation of such a sphere is \(x^2+y^2+z^2 - px - qy - rz = 0\). 3. The center of this sphere is \((\frac{p}{2}, \frac{q}{2}, \frac{r}{2})\). 4. The plane passes through the fixed point (a,b,c). Substitute this into the plane equation to get a condition on p, q, r. 5. Use the coordinates of the center to relate p, q, r to x, y, z of the locus. Substitute this into the condition from step 4 to find the locus equation.

Step 3: Detailed Explanation:

Let the variable plane cut the axes at \(A(p,0,0)\), \(B(0,q,0)\), and \(C(0,0,r)\). The equation of this plane in intercept form is: \[ \frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1 \] Since this plane passes through the fixed point \((a,b,c)\), these coordinates must satisfy the plane's equation: \[ \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 1 \quad (). \] Now consider the sphere OABC. The equation of a sphere passing through the origin is \(x^2+y^2+z^2+2ux+2vy+2wz=0\). Since it passes through A(p,0,0), B(0,q,0), and C(0,0,r), we find \(2u=-p, 2v=-q, 2w=-r\). The equation of the sphere is \(x^2+y^2+z^2 - px - qy - rz = 0\). The center of this sphere is \((-\frac{-p}{2}, -\frac{-q}{2}, -\frac{-r}{2}) = (\frac{p}{2}, \frac{q}{2}, \frac{r}{2})\). Let the coordinates of the center be \((x_c, y_c, z_c)\). To find its locus, we will just use \((x, y, z)\). \[ x = \frac{p}{2} \implies p = 2x \] \[ y = \frac{q}{2} \implies q = 2y \] \[ z = \frac{r}{2} \implies r = 2z \] Now, we substitute these expressions for p, q, and r into the condition we found earlier (\(\)): MATH_cae518a31add429abcfb8a7671f5ac05 Multiply the entire equation by 2: MATH_994e15a07100467eb432ac14b8191c2d This is the equation of the locus of the center of the sphere.
Step 4: Final Answer:
The locus of the centre of the sphere OABC is \(\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2\), which corresponds to option (B).
Was this answer helpful?
0