The equation of the auxiliary circle of the hyperbola $4 x^{2} - 9 y^{2} = 36$ is

x^{2} + y^{2} = 81

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x^{2} + y^{2} = 9

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x^{2} + y^{2} = 16

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

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Solution

The correct option is B $x^{2} + y^{2} = 9$
for hyperbola $a > b$
equation of auxiliary circle is $x^{2} + y^{2} = a^{2}$
$x^{2} + y^{2} = 9$ is the equation of auxiliary circle for the given hyperbola

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Similar questions

What the equation of the auxiliary circle of a hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

Equation of the hyperbola with length of the latusrectum $\frac{9}{2}$ and e = $\frac{5}{4}$ is

\frac{x^{2}}{64} - \frac{y^{2}}{36} = 1

x^{2} + y^{2} = 225

\frac{x^{2}}{144} + \frac{y^{2}}{81} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} + y^{2} = a^{2}

25 y^{2} + 250 y - 16 x^{2} - 32 x + 209 = 0

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