Question

If the two circles $x^2+y^2+2gx+c=0$ and $x^2+y^2-2fy-c=0$ have equal radius then locus of $(g,f)$ is

• A. $x^2+y^2-c^2$
• B. $x^2+y^2-2c$
• C. $x^2-y^2-2c$
• D. $x^2+y^2-2c^2$

Answer 1

Answer: (C)

$x^2-y^2-2c$

Given the equation of the circles $x^2+y^2+2gx+c=0$.....(1) and $x^2+y^2-2fy-c=0$ .......(2).

Now center of the first circle is $(-g,0)$ and radius is $\sqrt{g^2-c}$ and the centre of the second circle is $(0,f)$ and radius is $\sqrt{f^2+c}$.

Now according to the problem we've,

$\sqrt{g^2-c}=\sqrt{f^2+c}$

or, $f^2-g^2+2c=0$

or, $g^2-f^2-2c=0$.

So the locus of $(g,f)$ is $x^2-y^2-2c=0$.