Question

Find the equation of the sphere that passes throught a(1,2,-3), with center b (2,4,0).

Answer 1

The equation of sphere is (x -1 )^{2}  + (y - 2 )^{2}  + (z+3 )^{2}  = 14.

According to the statement

we have to find the equation of the sphere which passes through the a(1,2,-3), with center b (2,4,0).

So, For this purpose, we know that the

Sphere is In analytical geometry, if “r” is the radius, (x, y, z) is the locus of all points and $(x_{0} , y_{0} , z_{0})$

is the center of a sphere, then the equation of a sphere is given by: $(x -x_{0} )^{2} + (y - y_{0} )^{2} + (z-z_{0} )^{2} = r^{2}$.

So, Here from given equation

The sphere that passes through a(1,2,-3), with center b (2,4,0).

here the locus point is a(1,2,-3).

The radius of sphere is

$Radius = \sqrt{(1-2)^{2} + (2-4)^{2} + (-3-0)^{2} } \\Radius = \sqrt{(1 + 4 +9 }$

So, radius is square root 14.

And then the equation of sphere become

$(x -1 )^{2} + (y - 2 )^{2} + (z+3 )^{2} = 14$.

So, The equation of sphere is (x -1 )^{2}  + (y - 2 )^{2}  + (z+3 )^{2}  = 14.

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