The equation of a sphere centered at the point (3,5,3) and tangent to the plane z=0 isx2+px+y2+qy+z2+rz=sx2+px+y2+qy+z2+rz= SS = ?

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RizerMix · Accepted Answer

To solve this problem, we need to use the formula of a sphere and the equation of a plane. The equation of a sphere with center (a,b,c) and radius r is given by:

${(x - a)}^{2} + {(y - b)}^{2} + {(z - c)}^{2} = r^{2}$

In this problem, the sphere is centered at (3,5,3), so we have:

${(x - 3)}^{2} + {(y - 5)}^{2} + {(z - 3)}^{2} = r^{2}$

We also know that the sphere is tangent to the plane z=0. This means that the distance from the center of the sphere to the plane z=0 is equal to the radius of the sphere. The distance from a point (x,y,z) to the plane with equation Ax + By + Cz + D = 0 is given by:

$d = \frac{| A x + B y + C z + D |}{\sqrt{A^{2} + B^{2} + C^{2}}}$

So, the distance from the center of the sphere (3,5,3) to the plane z=0 is:

$d = \frac{| 0 (3) + 0 (5) + 1 (3) + 0 |}{\sqrt{0^{2} + 0^{2} + 1^{2}}} = 3$

Therefore, the radius of the sphere is also 3. We can now substitute this value of r into the equation of the sphere to get:

${(x - 3)}^{2} + {(y - 5)}^{2} + {(z - 3)}^{2} = 3^{2}$

Simplifying the equation, we get:

$x^{2} - 6 x + y^{2} - 10 y + z^{2} - 6 z + 25 = 0$

Comparing this equation with the equation of a sphere, we have:

p = -6
q = -10
r = -6
s = 1

The sum of the coefficients of the equation is:

s + p + q + r + 1 = 1 - 6 - 10 - 6 + 1 = -20

Therefore, SS = -20.

The equation of a sphere centered at the

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