Question #223817

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

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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