Question

Find the equation of the circle concentric with the circle $2x^2+2y^2+8x+10y-39=0$and having its area equal to $16\pi$ sq units.

Or

Find the equation of the hyperbola whose conjugate axis is 5 and the distance between foci is 13.

Answer 1

Given equation of circle is $2x^2+2y^2+8x+10y-39=0$

$\Rightarrow x^2+y^2+4x+5y-\frac{39}{2}=0[divideby2$

Then, coordinates of its centre are $(-2,-5/2)$

The required circle is concentric with the above circle

$\therefore$ its centre will be $(-2,-5/2)$

Let r be the radius of the required circle, then its area is $\pi r^2$

But area =$16\pi$

$\pi r^2=16\pi \Rightarrow r^2=16\Rightarrow r=4[\because$ r cannot be negative]

Now, the equation of the required circle having centre $(-2,-\frac{5}{2})$ and radius 4 is

$(x+2{)}^2+{(y+\frac{5}{2})}^2=4^2$

$\Rightarrow x^2+4x+4+y^2+5y+\frac{25}{4}=16$

$\therefore 4x^2+4y^2+16x+20y-23=0$

Or

Let 2a and 2b be the and lengths of transverse and conjugate axes respectively and 2c be the distance between two foci.

We know that the equation of the hyperbola, when centre is origin, is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

We have , $2b=5and2c=13\Rightarrow b=\frac{5}{2}andC=\frac{13}{2}$

$\because c^2=a^2+b^2$

$\therefore a^2=c^2-b^2$

$\Rightarrow a^2={\left(\frac{13}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2=\frac{169-25}{4}=\frac{144}{4}=36$

On substituting the values of $a^{2}and{b}^2$in Eq. (i) , we get

$\frac{x^2}{36}\frac{y^2}{25/4}=1\Rightarrow \frac{x^2}{36}-\frac{4y^2}{25}=1$

$\Rightarrow 25x^2-144y^2=900$

Hence , the equation of hyperbola is $25x^2-144y^2=900$