Question #243733

The differential equation of (x-a)^{2}+(y-b)^{2}+z^{2}=2019 where a, b are arbitrary constants is.

A) z^2(p^2+q^2)=1. B)z^2(p^2+q^2)=2019.

C) z^2(p^2+q^2+1)=1.

D) z^2(p^2+q^2+1)=2019

Expert's answer

Differentiating partially with respect to "x" and "y," we get

Squaring and adding these equations, we have

Since "(x-a)^2+(y-b)^2=z," we get

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