How can I show that a circle that is tangent to $x$-axis and $y$-axis has radius $R$ and center $(R,R)$ using algebra?
So I have this concept about it. Since the point on $y$ tangent to the circle (I'll call it $Y$) is $(0,Y)$ and on $x$ (I'll call it $X$) is $(X,0)$. Let's call the center $E$. From $X$ to $E$ and $Y$ to $E$ are the radii. So $XE=YE$. Then I'll use the distance formula (which I don't know how to render here) but I can't continue from it.
So, how can I continue?
$\endgroup$ 42 Answers
$\begingroup$The equation to a circle is $$(x-x_c)^2+(y-y_c)^2=r^2$$We know the circle is tangent to $x,y$ axi.
We also know the derivative of the circle equation (with respect to $x$): $$2x+2(y-y_c)y'=0$$$$y'=\frac x{y-y_c}$$$$y=\sqrt{r^2-(x-x_c)^2}+y_c$$$$y'=\frac x{\sqrt{r^2-(x-x_c)^2}}$$
Setting the denominator equal to $0$, we get $$0=\sqrt{r^2-(x-x_c)^2}$$$$x=\pm r+x_c$$For this to be our $y-intercept$, we have $x=0$.$$x_c=\pm r$$This puts our $x-coordinate$ for the center to actually be $\pm r$.
We can repeat the process by differentiating with respect to $y$ and setting the denominator equal to $0$ to get $y_c=\pm r$.
$\endgroup$ 4 $\begingroup$Consider the diagram below of a circle with its center in the first quadrant that is tangent to both coordinate axes:
Let $(x, y)$ be the center of a circle with radius $r$ (we do not assume the center of the circle is in the first quadrant). If the circle is tangent to the $y$-axis, then the point of tangency must be $(0, y)$ since the perpendicular to the $y$-axis must be horizontal. Since the distance from the center to the point of tangency is $r$, we obtain
\begin{align*}
|x - 0| & = r\\
|x| & = r\\
x & = \pm r
\end{align*}
Since the circle is tangent to the $y$-axis, the point of tangency must be $(x, 0)$ since the perpendicular to the $x$-axis must be vertical. Since the distance from the center to the point of tangency is $r$, we obtain
\begin{align*}
|y - 0| & = r\\
|y| & = r\\
y & = \pm r
\end{align*}
These equations lead to the four possible solutions $(\pm r, \pm r)$ for the center of the circle, as shown in the diagram below.
