Question

The equation of auxiliary circle of hyperbola $25y^2+250y-16x^2-32x+209=0$ is

• A. $(x+1{)}^2+(y+5{)}^2=25$
• B. $x^2+y^2=25$
• C. $x^2+y^2=16$
• D. $(x+1{)}^2+(y+5{)}^2=16$

Answer 1

The correct option is D $(x+1{)}^2+(y+5{)}^2=16$

$25y^2+250y-16x^2-32x+209=0$
$\Rightarrow 25(y^2+10y)-16(x^2+2x)+209=0$
Now completing the square, we get
$25(y+5{)}^2-16(x+1{)}^2=400$
$\Rightarrow$ $\frac{(y+5{)}^2}{16}-\frac{(x+1{)}^2}{25}=1$

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