Question

find the equation of sphere whose is centre(2,3,0) which passes through (1,0,2)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

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

Step-by-step explanation:

given that the centre of the sphere is O (2,3,0) and it passes through P(1,0,2)

=> OP is the radius of the sphere

so OP (radius) = $\sqrt{(2-1)^{2}+(3-0)^{2} +(0-2)^{2} }$

                        = $\sqrt{1^{2}+3^{2}+(-2)^{2} }$

                        = $\sqrt{1+9+4}$

                        r = $\sqrt{14}$

therefore the equation of the circle is given by,

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

ie.,  $(x-2)^{2} + (y-3)^{2} + (z-0)^{2} = \sqrt{14} ^2$

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