Question

If the two circles x2+y2+2gx+c = 0 and x2+y2–2fy-c = 0 have equal radius then locus of (g,f) is
1) x2 + y2 = c2 2) x2 - y2 = 2c 3)x - y2= c²
4) x2 + y2 = 2c2​

Answer 1

SoluTion:

Given the equation of the circles

$\bf{x²+y²+2gx+c = 0}$.....(1)

$\bf{x²+y²-2fy-c = 0}$

Now center of the first circle is $\textit {(−g,0)}$ and radius is $\bf{\sqrt{g²-c}}$ and the centre of the second circle is $\textit {(0,f)}and\; radius\; is \bf{\sqrt{f²+c}}$

According To Sum

➝ $\bf{\sqrt{g²-c} = \sqrt{f²-c}}$

➝ $\bf{g²-f²-c = 0 (or) f²-g²+c = 0}$

So The Locus is $\pink{\textit{x²-y²-2c = 0}}$

$\Large\red{\underline{\bf{Required\;AnSweR:}}}$

Option B $\pink{\sf{x²-y² = 2c}}$