Equation of The Circle Tangent to Each Axis and Passing Through a Point

Question

Find the equation of the circle touching each axis and passing through $(2,1)$

My Attempt:

Since the circle touches both axes, the x-coordinate of the center = y-coordinate of the center = radius. If $h$, $k$ and $r$ are the x-coordinate, y-coordinate and radius then $$h=k=r.$$

Answer 1

Let $(x_{0}, y_{0})$ denote the coordinates of the center of the circle. Then $x_{0}=y_{0}$. Also $r$ is $x_{0}$. Now, we have: $$(2-x_{0})^2 +(1-x_{0})^2 = x_{0}^2 \implies x_{0}^2 -6x_{0}+5=0$$

Answer 2

The equation of your circle is then $(x-r)^2+(y-r)^2=r^2$. Plug in $x=2,y=1$ and solve for $r$.

Answer 3

Let the radius of the desired circle be r. Since the circle passes through the point (2,1) and it is tangent to the coordinate axes, the center of the circle is in the first quadrant with positive coordinates of (r,r). Therefore the equation of the circle is (x-r)^2+(y-r)^2=r^2. plugging (2,1) for (x,y) and solving for r results in r=1 or r=5. Therefore we have two such circles.