The correct option is D (x+1)2+(y+5)2=16
25y2+250y−16x2−32x+209=0
⇒25(y2+10y)−16(x2+2x)+209=0
Now completing the square, we get
25(y+5)2−16(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