circlesgeometrysoddy-circles
We have three circles tangent to each other with radii $1$, $2$, and $3$. Another circle is tangent to the other circles; find the radius of that circle using elementary geometry, without the Soddy Circles formula (aka, Descartes' Theorem).
My Attempt: I tried to draw a triangle with points $A$, $B$, and $C$, and then I proved that one of its angles is $90^\circ$.
Let's get a figure in here, eh?
We can exploit the fact that the points of mutual tangency of the three inner circles (red points in the figure) form an equilateral triangle; we also know that the red points are the midpoints of the segments formed by joining any two of the three blue points (the centers of the inner circles).
We then deduce that the triangle formed by the red points is half the scale of of the triangle formed by the blue points, and find that the equilateral triangle formed by the blue points has a side length of 6 ft., and that the inner circles have a radius of 3 ft. Using the law of cosines, we reckon that the distance from a blue point to the center of the triangle formed by the blue points is $2\sqrt{3}$ ft. Adding to that the radius of an inner circle, we find that the radius of the outer circle is $3+2\sqrt{3}$ ft.
The area of the outer circle is $\approx$ 131.27 square feet.
If circles of radius $r_1$, $r_2$, $r_3$, $r_4$ are mutually tangent, and also circles of radius (say) $r_2$, $r_3$, $r_4$, $r_5$, then we have
$$\begin{align} \left( k_1 + k_2 + k_3 + k_4 \right)^2 &= 2 \left( k_1^2 + k_2^2 + k_3^2 + k_4^2 \right) \\ \left( k_2 + k_3 + k_4 + k_5 \right)^2 &= 2 \left( k_2^2 + k_3^2 + k_4^2 + k_5^2 \right) \end{align}$$ with $k_i = \pm 1/r_i$ as desired.
If you need the fourth circle's radius, then, of course, you can solve the equations in stages: get $k_4$ from the first, and use that in the second to get $k_5$.
If you don't really care about the fourth radius, we can eliminate $k_4$ from the equations. (For instance, subtract one from the other to get rid of the $k_4^2$ terms, and solve for $k_4$ in the resulting linear equation. Then substitute this $k_4$ into either of the original equations.) Assuming $k_3 \neq k_5$, we arrive at this relation: $$16 k_1^2 + 16 k_2^2 + k_3^2 + k_5^2 + 16 k_1 k_2 - 8 k_1 k_3 - 8 k_1 k_5 - 8 k_2 k_3 - 8 k_2 k_5 - 2 k_3 k_5 = 0$$ Thus, $$k_5 = 4 k_1 + 4 k_2 + k_3 \pm 4 \sqrt{k_1 k_2 + k_2 k_3 +k_3 k_1}$$
Best Answer
As you have observed, $AB$ and $AC$ are perpendicular, so take $A$ as the origin and assign coordinates $B(4,0)$ and $C(0,3)$.
Let the centre of the inner circle be at $(a,b)$ and the radius is $r$. Then, considering distances, the following equations apply:
$$a^2+b^2=(1+r)^2$$ $$(a-4)^2+b^2=(3+r)^2$$ $$a^2+(b-3)^2=(2+r)^2$$
Subtracting the first equation from the second gives rise to $$a=\frac{2-r}{2}$$ Subtracting the first equation from the third gives rise to $$b=\frac{3-r}{3}$$
Substituting these into the first equation then leads to the quadratic equation $$23r^2+132r-36=0$$
The roots are $$r=\frac{6}{23}$$ which is the answer you are looking for, and $$-6$$ which corresponds to the radius $6$ of the outer kissing circle. All of this can be verified by applying Descartes' Formula.