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Mathematics
The equation of auxilary circle of hyperbola (x2/9)-(y2/16)=1
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Q. The equation of auxilary circle of hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$
Conic Sections
A
$x^2+y^2=9$
B
$x^2+y^2=16$
C
$x^2+y^2=41$
D
$x^2+y^2=7$
Solution:
Equation of auxilary circle
$\Rightarrow x^2+y^2=9$