Explanation:
C Given,
\(x^2+y^2+4 x-4 y+4=0\)
\(\qquad \frac{d}{d x}\left(x^2+y^2+4 x-4 y+4\right)=0\)
\(2 x+2 y \frac{d y}{d x}+4-4 \frac{d y}{d x}=0\)
\(\frac{d y}{d x}(2 y-4)=-(4-2 x)\)
\(\frac{d y}{d x}(4-2 y)=4+2 x\)
\(\frac{d y}{d x}(2-y)=x+2\)
\(\frac{d y}{d x}=\frac{x+2}{2-y}\)
If circle touches \(x\) - axis.
\(\frac{d y}{d x}=0\)
\(x+2=0\)
\(x=-2\)
\(x^2+y^2+4 x-4 y+4=0\)
\((-2)^2+y^2+4(-2)-4 y+4=0\)
\(y^2-4 y-8+8=0\)
\(y^2-4 y=0\)
\(y(y-4)=0\)
\(y=0 \text { or } y=4\)
So, \((-2,0) \&(-2,4)\)
If circle touches \(y\)-axis,
\(2-y=0\)
\(y=2\)
\(x^2+y^2+4 x-4 y+4=0\)
\(x^2+(2)^2+4 x-4(2)+4=0\)
\(x^2+4 x+8-8=0\)
\(x^2+4 x=0\)
\(x(x+4)=0\)
\(x=0 \text { or } x=-4\)
So, \((0,2) \&(-4,2)\)
Hence, circle touches both \(\mathrm{x} \& \mathrm{y}\) - axis.