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​

Answer 1

Given :

$C_1 = x^2+y^2+2gx+c=0 \\\\C_2 = x^2+y^2-2fy-c=0$

To find : locus (g,f)

solution :

consider the first circle

$C_1 = x^2+y^2+2gx+c=0$

here ,

center =  $\frac{-\text{coff . of x }}{2}$

center  = (-g,0)

then radius = $\sqrt{g^2-c}$

considering the second circle

$C_2 = x^2+y^2-2fy-c=0$

here ,

center =  $\frac{-\text{coff . of y }}{2}$

C = (0,f)

radius =

$\sqrt{f^2-(-c)}\\\\\sqrt{f^2+c}$

according to the question : radius are equal

thus

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

squaring both sides

g² -c = f²+c

g² -f² = 2c

replacing g by x and f by y we get

locus of (g,f) is

x²-y² = 2c

hence ,The locus of (g ,f) is​ x²-y² = 2c