Then equation of auxiliary circle of the ellipse $16 x^{2} + 25 y^{2} + 32 x - 100 y = 284$ is

x^{2} + y^{2} + 2 x - 4 y - 20 = 0

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x^{2} + y^{2} + 2 x - 4 y = 0

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(x + 1)^{2} + (y - 2)^{2} = 400

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(x + 1)^{2} + (y - 2)^{2} = 225

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Solution

The correct option is B $x^{2} + y^{2} + 2 x - 4 y - 20 = 0$
Given ellipse is $16 x^{2} + 25 y^{2} + 32 x - 100 y = 284$
$\Rightarrow (16 x^{2} + 32 x) + (25 y^{2} - 100 y) = 284$
$\Rightarrow 16 (x^{2} + 2 x) + 25 (y^{2} - 4 y) = 284$
$\Rightarrow 16 (x^{2} + 2 x + 1 - 1) + 25 (y^{2} - 4 y + 4 - 4) = 284$
$\Rightarrow 16 (x + 1)^{2} + 25 (y - 2)^{2} = 284 + 16 + 100 = 400$
$\Rightarrow \frac{(x + 1)^{2}}{25} + \frac{(y - 2)^{2}}{16} = 1$

Hence auxiliary circle of above ellipse is given by,
$(x + 1)^{2} + (y - 2)^{2} = 25$
$\Rightarrow x^{2} + y^{2} + 2 x - 4 y - 20 = 0$

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