Question

Find the equation of a circle concentric with the circle $2x^2+2y^2-6x+8y+1=0$ and of double its area.

Answer 1

Center of the given circle

$x^2+y^2-3x+4y+\frac{1}{2}=0$

is $(\frac{3}{2},-2)$

Area of the given cirle $=\pi r_1^2$

Area of circle to be found $=2\pi r_1^2$

$\Rightarrow r_2=\sqrt{2}{r}_1$

$g_2=g_1=\frac{-3}{2}$

$f_2=f_1=2$

$r_1=\sqrt{g_1^2+f_1^2-c}$

$=\sqrt{\frac{9}{4}+4-\frac{1}{2}}$

$\sqrt{\frac{9+16-2}{4}}$

$=\sqrt{\frac{23}{4}}$

$r_2=\sqrt{\frac{23}{2}}$

and $r_2=\sqrt{g_2^2+f_2^2-c_2}$

$=\sqrt{\frac{9}{4}+4-c_2}=\sqrt{\frac{23}{2}}$

$\Rightarrow \frac{25}{4}-c_2=\frac{23}{2}$

$\Rightarrow c_2=\frac{25}{4}-\frac{23}{2}=\frac{-21}{4}$

$\therefore$ Equation of required circle:

$x^2+y^2-3x+4y-\frac{21}{4}=0$

$\Rightarrow 4x^2+4y^2-12x+16y-21=0$