Question

A cirde, centre (3 3.0) and radius 1, is drawn in the plane : =. Fall
equation of the sphere which passes through this circle and through the
(1, 1, 1).​

Answer 1

Question :-

A circle, centre (2, 3, 0) and radius 1, is drawn in the plane z = 0. Find the equation of the sphere which passes through this circle and through the point (1, 1, 1).

$\large\underline\purple{\bold{Solution :- }}$

The equation of circle passes through the point (3, 3, 0) having radius 1 is given by

$\rm :\implies\: {(x - 2)}^{2} + {(y - 3)}^{2} + {z}^{2} = 1$

Now,

Given

• The equation of plane, z = 0.

So,

The equation of sphere which passes through this circle is given by

$\rm\: {(x - 2)}^{2} + {(y - 3)}^{2} + {z}^{2} - 1 + kz = 0 - - (i)$

Since, (i) passes through (1, 1, 1), so

$\rm :\implies\: {(1 - 2)}^{2} + {(1 - 3)}^{2} + {(1)}^{2} - 1 + k(1) = 0$

$\rm :\implies\:1 + 4 + 1 - 1 + k = 0$

$\rm :\implies\:k \: = \: - \: 5$

On substituting k = - 5, in equation (i), we get

$\rm :\implies\: {(x - 2)}^{2} + {(y - 3)}^{2} + {z}^{2} - 1 - 5z = 0$

$\rm :\implies\: {x}^{2} + 4 - 4x + {y}^{2} + 9 - 6y + {z}^{2} - 5z - 1 = 0$

$\rm :\implies\: {x}^{2} + {y}^{2} + {z}^{2} - 4x - 6y - 5z + 12 = 0$

Tʜᴜs,

• The required equation of the sphere is

$\bigstar \: \: { \large \boxed{ \red{ \tt \: {x}^{2} + {y}^{2} + {z}^{2} - 4x - 6y - 5z + 12 = 0}}}$

Answer 2

hope you understand

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(prompto)