HINT:
If the coordinate of the center is $(-b,-5)$ as it lies on $y=-5$
Observe that $$\tan 22.5^\circ=-\frac5b$$
Consider the point-slope formula:
$$\left( y-y_1 \right)=m\left( x-x_1 \right).$$
This is nothing more than a consequence of the definition of slope. More specifically,
$$m=\frac{\left( y_2-y_1 \right)}{\left( x_2-x_1 \right)}\Rightarrow \left( y_2-y_1 \right)=m\left( x_2-x_1 \right).$$
If we are in a situation where we already know the slope $m$ and we already know a point $( x,y )$, we can drop one of the points in that last form (the $_2$ in $x_2$ and $y_2$ for example) and write the point-slope formula as above.
Using the point-slope form
$$(y−y_1)=m(x−x_1),$$
with $m=\frac{3}{4}$, and the point $(1,-3)$ we substitute our given $m$, $x$, and $y$ to get.
$$(y−(−3))=\frac{3}{4}(x−1).$$
If you want to convert this to the slope intercept form then the process is
\begin{align*}
&(y−(−3))=\frac{3}{4}(x−1) \\
\Rightarrow & y+3=\frac{3}{4}x-\frac{3}{4} \\
\Rightarrow & y=\frac{3}{4}x-\frac{3}{4}-3 \\
\Rightarrow & y=\frac{3}{4}x-\frac{15}{4}.
\end{align*}
Note that I still recommend my original solution where we start with the slope-intercept form. It is much faster (and seems much more sensible to me), although I have noticed a strange reluctance to demonstrate that method directly in the textbook (Bittinger) that I am currently using for my classes.
Best Answer
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 $y_c=\pm r$.