Question

The equation of the circle concentric to the circle $2x^2+2y^2-3x+6y+2=0$ and having area double the area of this circle, is

• A. $8x^2+8y^2-24x+48y-13=0$
• B. $16x^2+16y^2+24x-48y-13=0$
• C. $16x^2+16y^2-24x+48y-13=0$
• D. $8x^2+8y^2+24x-48y-13=0$

Answer 1

Answer: (C)

The equation of given circle can be written as
$x^2+y^2-\frac{3}{2}x+3y+1=0$
whose centre is $\left(\frac{3}{4},-\frac{3}{2}\right)$
and radius, $r=\sqrt{\frac{9}{16}+\frac{9}{4}-1}$
$=\sqrt{\frac{9+36-16}{16}}$
$=\sqrt{\frac{29}{16}}$
$\therefore$ Area of circle $=\pi r^2$
$=\pi \left(\frac{29}{16}\right)=\frac{29\pi }{16}$
$\Rightarrow$ Area of required circle $=2\times \frac{29\pi }{16}$
$=\frac{29\pi }{8}$
Let $R$ be the radius of required circle.
$\therefore R^2=\frac{29}{8}.$
Now, equation of circle is
${\left(x-\frac{3}{4}\right)}^2+{\left(y+\frac{3}{2}\right)}^2=\frac{29}{8}$.
$\Rightarrow x^2-\frac{3x}{2}+\frac{9}{16}+y^2+3y+\frac{9}{4}-\frac{29}{8}=0$
$\Rightarrow 16x^2+16y^2-24x+48y-13=0$.