Question

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

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

Answer 1

Answer: (C)

$16{{x}^{2}}+16{{y}^{2}}-24x+48y-13=0$

Answer 2

The equation of given circle can be written as
$x^{2}+y^{2}-\frac{3}{2} x+3 y+1=0$
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
$\left(x-\frac{3}{4}\right)^{2}+\left(y+\frac{3}{2}\right)^{2}=\frac{29}{8}$.
$\Rightarrow x^{2}-\frac{3 x}{2}+\frac{9}{16}+y^{2}+3 y+\frac{9}{4}-\frac{29}{8}=0$
$\Rightarrow 16 x^{2}+16 y^{2}-24 x+48 y-13=0$.