Find radius of a circle which is tangent to three known lines

Question

I need to find the equation for a circle which is tangent to the following three lines:

y=0

x=0

y=-x+0.338334

For the last tangent line equation, I know that it is tangent at the point (0.169167, 0.169167) However, for the other two I do not know the exact point of tangency, only that the circle is tangent to the x and y axis.

Note: Picture is not to scale, only meant to be a reference image

Answer 1

The point must have equal coordinates because the distances to the x-axis and the y-axis must coincide.

So P(u/u)

The distance from the point (0.169167/0.169167) must be u.

So

$$\sqrt{2}(u-0.169167)=u$$

which gives

$$u=\frac{\sqrt{2}*0.169167}{\sqrt{2}-1} = 0.577572$$

This u is the radius of the desired circle.

Answer 2

Let us consider two similar right angled triangles whose angles will be the same, one of them the larger one.

We form the larger right angled triangle by drawing a line from the origin of the graph to the centre of the circle, a line from the centre of the circle to the y-axis tangent, and finally a line from the y-axis tangent to the origin.

We form the smaller right angled triangle by drawing a line from the origin of the graph to the tangent of the negative slope, a line from the negative slope tangent to the y-axis to make a right angle, and finally a line from that y-axis point back to the origin.

Let $r$ be the radius of the circle

Using trigonometry (sine of angle made by hypotenuse to y-axis) for the larger and smaller triangles respectively we have,

$\large \frac{r}{r+\sqrt{0.169167^2+0.169167^2}}=\frac{0.169167}{\sqrt{0.169167^2+0.169167^2}}$

This leads to

$\large r=\frac{0.169167\sqrt{0.169167^2+0.169167^2}}{\sqrt{0.169167^2+0.169167^2}-0.169167}=0.577572$

Answer 3

The given circle is defined by $$ C:=\{(x,y) \in \mathbb{R}^2|\quad (x-r)^2+(y-r)^2=r^2\}, $$ where $r>a:=0.338334$.

Since $C$ is tangent to the line $$ L=\{(x,y)\in \mathbb{R}^2:\ y=-x+a\}, $$ then $r$ is precisely the distance from the center $(r,r)$ of $C$ to the line $L$, i.e. $$ r=\frac{|r+r-a|}{\sqrt{1+1}}=\frac{|2r-a|}{\sqrt{2}}=\sqrt{2}r-\frac{a}{\sqrt{2}}. $$ Hence $$ r=\frac{a}{\sqrt{2}(\sqrt{2}-1)}=\left(1+\frac{1}{\sqrt{2}}\right)a \approx 0.5775722657 $$