If the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ touches $x-$ axis then \[\]
(a)$g=f$\[\]
(b)${{g}^{2}}=c$\[\]
(c)${{f}^{2}}=c$\[\]
(d)${{g}^{2}}+{{f}^{2}}=c$\[\]
VerifiedHint: We use the fact that a circle touching the $x-$axis will have the absolute value of $y-$coordinate of the centre equal to the length of the radius. We equate the radius $r$ of the given general circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and find a relation between $r,f$ and then simplify.
Complete step by step answer:
We know from the general second degree equation of circle in plane in two variables with real constants $a,b,g,f,c$ is given by
\[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\]
We also know that the radius of the above circle is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ and centre is given by $\left( -g,-f \right)$.
Let the radius of the given circle be $r$. So we have
\[r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]
We square both sides to have
\[{{r}^{2}}={{g}^{2}}+{{f}^{2}}-c.....\left( 1 \right)\]
We know that the absolute value of $x-$coordinate is the distance of a point from $y-$axis and absolute value of $y-$coordinate is the distance of a point from $x-$axis. So the distance of the centre $\left( -g,-f \right)$ from the $x-$axis is the absolute value of the $y-$coordinate that is $-f$. Since the circle touches the $x-$axis the radius will be equal to distance from the centre to the tangent $x-$axis which is equal to absolute value of $y-$coordinate. So we have;
\[r=\left| -f \right|\]
We square both sides of above equation to have;
\[{{r}^{2}}={{f}^{2}}\]
We put the above obtained result in equation (1) to have;
\[\begin{align}
& {{f}^{2}}={{g}^{2}}+{{f}^{2}}-c \\
& \Rightarrow 0={{g}^{2}}-c \\
& \Rightarrow {{g}^{2}}=c \\
\end{align}\]
So, the correct answer is “Option B”.
Note: We note that similarly a circle touching the $y-$axis will have the absolute value of $x-$coordinate of the centre equal to the length of the radius. . We can alternatively use the perpendicular distance between a point $\left( {{x}_{1}},{{y}_{1}} \right)$ from a line $ax+by+c=0$ is given by $d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ taking the point $\left( -g,-f \right)$ and the equation of $x-$axis that is $y=0$.We also note that tangent is perpendicular radius of the circle.
