Question

If the radius of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ be $r$, then it will touch both the axes, if \[\]
A.$g=f=c$\[\]
B.$g=f=c=r$\[\]
C.$g=f=\sqrt{c}=r$\[\]
D.$g=f$ and ${{c}^{2}}=r$\[\]

Answer 1

Hint: We use the fact that a circle touching $x-$axis will have the absolute value of $y-$coordinate of the centre equal to the length of the radius and similarly a circle touching $y-$axis will have the absolute value of $x-$coordinate of the centre equal to the length of the radius. We recall how to the centre of a general equation of a circle is $\left( -g,-f \right)$ and the radius is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. We find a relation among $g,f,c$. \[\]

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 coordinates of centre is given by $\left( -g,-f \right)$.
We are given 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 $x-$axis is the absolute value of the $y-$coordinate that is $-f$and the distance of the centre $\left( -g,-f \right)$ from $y-$axis is the absolute value of the $x-$coordinate that is $-g$. Since the circle touches both $x-$axis and $y-$axis, the radius will be equal to distance from the centre to the tangent $x-$axis and $y-$axis. So we have;
\[r=\left| -f \right|,r=\left| -g \right|\]
 We square both sides above two equation to have;
\[\begin{align}
  & {{r}^{2}}={{f}^{2}},{{r}^{2}}={{g}^{2}} \\
 & \Rightarrow {{r}^{2}}={{f}^{2}}={{g}^{2}} \\
\end{align}\]
We take square root to have;
\[r=f=g.........\left( 2 \right)\]
 We put the above relation in equation (1) to have;
\[\begin{align}
  & {{r}^{2}}={{r}^{2}}+{{r}^{2}}-c \\
 & \Rightarrow {{r}^{2}}=c \\
\end{align}\]
We take square root both sides to have;
\[\Rightarrow r=\sqrt{c}......\left( 3 \right)\]
So we have from (2) and (3) to have;
\[g=f=\sqrt{c}=r\]

So, the correct answer is “Option C”.

Note: We can alternatively use the perpendicular distance between a point $\left( {{x}_{1}},{{y}_{1}} \right)$ from a line $ax+by+c=0$ which 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 $y=0$ and then equation of $y-$axis that is $x=0$. We note that tangent is always perpendicular to the radius of the circle.