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A sphere of constant radius r passes through origin and cuts the axes in A,B,C .find the locus of the foot perpendicular from the origin to the plane ABC.
let co-ordinates of A,B,C be respectively (a,0,0),(b,0,0),(c,0,0).Then the equation of plane ABC isx/a+y/b+z/c=1.... [Intercept form]Also equation of sphere OABC isx^2+y^2+z^2-ax-by-cz=0.....[Intercept form ]Now, centre of sphere OABC is (a/2,b/2,c/2) and radius is √a^2/4+b^2/4+c^2/4.By given condition,√a^2/4+b^2/4+c^2/4=ra^2+b^2+c^2=4r^2. .....(1)If F(x1,y1,z1) is the foot of perpendicular from the origin to the plane ABC,then OF is parallel to normal to plane ABC.Therefore,x1-0/(1/a)=y1-0/(1/b)=z1-0/(1/c)x1a=y1b=z1c=k(say)a=k/x1,b=k/y1,c=k/z1. Now,F(x1,y1,z1) lies on plane ABC Therefore, x1/a+y1/b+z1/c=1....(2)Putting values of a,b and c in (1) and (2), we get k^2(1/x1^2+1/y1^2+1/z1^2=4r^2and 1/k(x1^2+y1^2+z1^2)=1Eliminating k from these,we get(x1^2+y1^2+z1^2)^2 (x1^-2+y1^-2+z1^-2)=4r^2.Hence locus of F(x1,y1,z1) is (x^2+y^2+z^2)^2 (x^-2+y^-2+z^-2)=4r^2
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