Question

If the circle x2+y2+2x+fy+k=0 touches both the axes, find f and k.

Answer 1

\[(x-h)^2+(y-g)^2-r^2=0\Rightarrow x^2-2hx+h^2+y^2-2gy+g^2-r^2=0\]
\[\Rightarrow -2h=2 \Rightarrow h=-1\Rightarrow h^2=1\]
\[\Rightarrow -2g=f \Rightarrow g=-\frac{f}{2}\Rightarrow g^2=\frac{f^2}{4}\]
\[\Rightarrow k=h^2+g^2-r^2=1+\frac{f^2}{4}-r^2\]\
Since the circle touches both axes, the radius has to be 1. this means that
\[f=\pm2\text{ and }k=1\]

Answer 2

the reason the radius has to be 1 is because h=-1 so when x = -1, the circle has to touch the x axis, making y = 0 => (-1+1)^2+(0-g)^2=1^2 so g^2 = 1 and g = +/-1