The equation of the auxiliary circle of the hyperbola x264−y236=1 is
A
x2+y2=100
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B
x2+y2=28
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C
x2+y2=64
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D
x2+y2=36
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Solution
The correct option is Cx2+y2=64 For any hyperbola of the form x2a2−y2b2=1,
the auxiliary circle is x2+y2=a2
Now for hyperbolax264−y236=1,a=8andb=6
Thus the auxiliary circle is x2+y2=64