Find the radii of the two circles which pass through the point $(16,2)$ and touch both axes How can I find the radii of the two circles which pass through the point $(16,2)$ and touch both the axes? I've only ever seen demonstrations using three normal co-ordinates; or two normal co-ordinates and touching 1 axis. 
I tried something along the lines of:
$$
x^2 + y^2 + 2gx+2fx+c=0 \\
260 + 32g+4f + c = 0
$$
And since we know it also meets the co-ordinates $(0,y)$ and $(x,0)$:
$$
x^2 + 2gx = -c = y^2+2fy
$$
But I'm not sure if this is a valid approach. Plugging the latter into the former doesn't seem to go anywhere...
 A: Hint:
Circles which touch both axes in the first quadrant would be of form $(x-a)^2+(y-a)^2=a^2$ (why?).  You want this to pass through $(16, 2)$, so that helps find possible values of $a$.
A: Write the equation of the circle as $(x-a)^2+(y-b)^2=r^2$ so that the circle has centre $(a,b)$ and radius $r$.
Now if the circle touches the $x$-axis the centre must be distance $r$ from the axis, so that $b=\pm r$. Also $a=\pm r$ so the equation becomes $$(x\pm r)^2+(y\pm r)^2=r^2$$ or $$x^2+y^2\pm 2xr\pm 2yr+r^2=0$$
Substituting the values for $x=16, y=2$ into the equation gives one of four quadratics for $r$ (choices of sign) - three of which are obviously impossible, since the point is in the first quadrant and a circle which touches both axes is confined to a single quadrant.  Solve for $r$.
A: The Answer provided by Macavity is short and perfect. But I just want to tell the facts about a circle touches X axis or Y axis.
(i) If a circle touch X axis, its radius is the same value of Y-coordinate of the center. (That is r=abs(Y-coordinate )).
(ii) If a circle touch Y axis, its radius is the same value of X-coordinate of the center. (That is r=abs(X-coordinate ) ).
(iii) If a circle touch both axis's, The center will have X-coordinate=Y-coordinate = +/- Radius  
A: As the circle grows, its center moves along line x = y.
So, both $ (16,2) $ and  its reflected point $ (2,16) $ satisfy your last equation:  $ x^2 + 2gx =  y^2+2fy . $
Plug them in, solve for f and g.
