Three Dimensional Geometry Question 34
Question: A plane passes through a fixed point $ (p,q,r) $ and cut the axes in A,B,C. Then the locus of the centre of the sphere $ OABC $ is
Options:
A) $ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2 $
B) $ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=1 $
C) $ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=3 $
D) None of these
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Answer:
Correct Answer: A
Solution:
Let the co-ordinates of A, B and C be (a,0,0), (0,b,0) and (0,0,c) respectively. The equation of the plane is $ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $ Also, it passes through (p, q, r). So, $ \frac{p}{a}+\frac{q}{b}+\frac{z}{c}=1 $ Also equation of sphere passes through A, B, C will be $ x^{2}+y^{2}+z^{2}-ax-by-cz=0 $ If its centre $ (x_1,y_1,z_1) $ , then $ x_1=\frac{a}{2},y_1=\frac{b}{2},z_1=\frac{c}{2} $ \ $ a=2x_1,b=2y_1,c=2z_1 $ \ Locus of centre of sphere $ \frac{p}{x}+\frac{q}{y}+\frac{r}{z}=2 $ .
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