A spherical ball is kept at the corner of a r | vedclass

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(A) Let the radius of the sphere be $r$. Since the sphere touches the two perpendicular walls and the floor,we can set up a coordinate system where th

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$13$

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(A) Let the radius of the sphere be $r$. Since the sphere touches the two perpendicular walls and the floor,we can set up a coordinate system where the walls and floor are the coordinate planes $x=0, y=0, z=0$. The center of the sphere is at $(r, r, r)$.The equation of the sphere is $(x-r)^2 + (y-r)^2 + (z-r)^2 = r^2$.$A$ point on the sphere is given at distances $9, 16, 25$ from the walls and floor,so the coordinates of this point are $(9, 16, 25)$.Substituting this point into the sphere's equation:$(9-r)^2 + (16-r)^2 + (25-r)^2 = r^2$Expanding the terms:$(81 - 18r + r^2) + (256 - 32r + r^2) + (625 - 50r + r^2) = r^2$$3r^2 - 100r + 962 = r^2$$2r^2 - 100r + 962 = 0$Dividing by $2$:$r^2 - 50r + 481 = 0$Factoring the quadratic equation:$(r-13)(r-37) = 0$Thus,$r = 13$ or $r = 37$.The possible radius is $13$.

$A$ spherical ball is kept at the corner of a rectangular room such that the ball touches two perpendicular walls and lies on the floor. If a point on the sphere is at distances of $9, 16, 25$ from the two walls and the floor,then a possible radius of the sphere is

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