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Question

The equation of auxiliary circle of hyperbola 25y2+250y16x232x+209=0 is

A
(x+1)2+(y+5)2=25
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B
x2+y2=25
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C
x2+y2=16
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D
(x+1)2+(y+5)2=16
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Solution

The correct option is D (x+1)2+(y+5)2=16
25y2+250y16x232x+209=0
25(y2+10y)16(x2+2x)+209=0
Now completing the square, we get
25(y+5)216(x+1)2=400
(y+5)216(x+1)225=1

Now the auxiliary circle has the center same as that of hyperbola and diameter is equal to the length of the transverse axis.
The equation of auxiliary circle is
(x+1)2+(y+5)2=16

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