Finding the equation of 4 circles given 3 tangents, one of which is oblique

Question

The question is to find the equation of the four circles tangent to the x-axis, the y-axis and the line $x+y=2$. I have drawn out a diagram and have identified the 4 circles but I am stuck on how to find their equations.

Answer 1

There are many ways to approach this. For example, if you know about incircle and excircles of triangles, then just follow that construction or other known facts (such as: the point of tangency of the excircle $\Gamma_A$ are distance $s=\frac12(a+b+c)$ from $A$, the in-/ex-centres are on the angle bisector, the inradius $r=\Delta/s$, exradius $r_A=\Delta/(s-a)$, etc.).

However, you can also do it purely algebraically: A circle $x^2+y^2+Dx+Ey+F=0$ is tangent to $Ax+By+C=0$ if and only if solving the simultaneous equation gives repeated roots. Eliminating $x$ (assuming $A\neq 0$, otherwise change $x\leftrightarrow y$) gives $$ (A^2+B^2) y^2 + (A^2 E-ABD+2BC) y + A^2 F - A C D + C^2=0 $$ and we want the discriminant to be zero: $$ (A^2 E-ABD+2BC)^2-4(A^2+B^2)(A^2 F - A C D + C^2)=0. $$ Expanding, you end up with $$\require{cancel} A^2 E^2 - 4 A^2 F - 2 A B D E + 4 A C D + B^2 D^2 - 4 B^2 F + 4 B C E - 4 C^2 = 0 $$ (remember we assumed $A\neq 0$). So the four circles satisfy \begin{align*} D^2 - 4 F&=0 &&x\text{-axis}\colon y=0\\ E^2 - 4 F&=0 &&y\text{-axis}\colon x=0\\ \cancel{E^2 - 4 F} - 2 D E - 8 D + \cancel{D^2 - 4 F} - 8 E - 16&=0 &&x+y-2=0.\\ \end{align*} Id est, we want to solve \begin{align*} D^2 - 4 F&=0\\ E^2 - 4 F&=0\\ D E + 4 D + 4 E + 8&=0.\\ \end{align*} From the first two equations, we have $D^2=E^2=4F$, so $D=\pm E$. Substituting into the third we have $D^2+8D+8=0$ (case $D=E$), or $-D^2+8=0$ (case $D=-E$). So we have $D=E=-4\pm2\sqrt2$ or $D=-E=\pm2\sqrt2$, hence $F=\frac14D^2=\dots$.

Answer 2

Unless I misunderstand the statement, there is one circle inside a triangle limited below by the $x$ axis, to the left by $y$ axis and by $y=2-x$ on the first quadrant (inside a triangle); there is another larger circle to the right limited on the left by the $y$ axis at the bottom by the $x$ axis and it is tangent to $y=2-x$ and the previous circle; another circle is on the 4th quadrant limited above by $x$ axis to the left by $y$ axis and it is tangent, say "north-east" to $y=2-x$. The last circle is on the 2nd quadrant and it is a obtained by the mirror image of the third circle an the line $y=x$. The centers of the circles are $C_1=(r_1,r_1)$, $C_2=(r_2,r_2)$, $C_3=(r_3,-r_3)$ and $C_4=(-r_4,r_4)$, where $r_i$ stands for the radius of the circle $i$. Notice that $r_3=r_4$.

It seems that you already know how to get the equation of the circles from the points of tangency. So, I won't discuss that.

One of the tangents for the first circle is determined by the intersection of the lines $y=x$ and $y=2-x$ at (1,1). The other tangents are $(r_1,0)$ and $(0,r_1)$, but you can get the radius by computing the distance to the first tangent point to the origin and dividing by 2.

For the second circle use the construction (Pythagoras) $r²_2+r²_2= \sqrt{1+1}+r_2$, where the square root comes from the distance of the tangent point along $y=x$ with the first circle. You can get a quadratic equation with only one solution satisfying the constraints. The other tangent points are $(r_2,0)$ and $(0,r_2)$.

For the third circle it is enough to notice that its center is along the line $y=-x$ which is parallel to $y=2-x$. The distance between these two lines is $\sqrt{2}$ (and that may be the radius). One of the tangent points is located at $x_{tan}=r_3+\sqrt{2} \: \cos(45)$ and $y_{tan}=-r_3+\sqrt{2}\: \sin(45)$, where the angle is measured counterclockwise from a horizontal line crossing the center of the circle. The other tangent points are $(r_3,0)$ and $(0,-r_3)$.

The last circle can be solved "mutatis mutandis" from the expression for the third one.