For the hyperbola x249−y225=1, the equation of auxillary circle is
A
x2+y2=49
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B
x2+y2=25
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C
x2+y2=10
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D
x2+y2=10074
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Solution
The correct option is Dx2+y2=49 For any Hyperbola of the form x2a2−y2b2=1,
The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of 2a, equal to the length of major axis and center same as center of Hyperbola.
Hence equation of Auxiliary circle of any standard Hyperbola will be x2+y2=a2
Here the given hyperbola is x249−y225=1, which is similar to standard form of the hyperbola.
Here a=7 and b=5
The equation of Auxiliary circle for the given hyperbola will also be x2+y2=(7)2