Let $ABCD$ be a rectangle with: $AB=a$, $BC=b$. Inside the rectangle we have to exteriorly tangents circles such that one is tangent to the sides $AB$ and $AD$,the other is tangent to the sides $CB$ and $CD$.
1. Find the distance between the centers of the circles(using $a$ and $b$).
2. When the radiums of both circles change the tangency point between both of them changes, and describes a locus. Find that locus.
Let the radius of the lower left circle be $r_1$, that of the upper circle be $r_2$, the horizontal component of the distance between the centers be $x$, and the vertical component of the distance between the centers be $y$. Also, letting $S=r_1+r_2$, we have the equations
\begin{eqnarray*} x&=&a-S\\ y&=&b-S\\ x^2+y^2&=&S^2 \end{eqnarray*}
Substituting the values from the first and second equations into the third, $(a-S)^2+(b-S)^2=S^2\Rightarrow S^2-2(a+b)S+(a^2+b^2)=0$. The quadratic formula gives $S=a+b\pm \sqrt{2ab}=\left( \sqrt{a}\pm \sqrt{b}\right)^2$, and since $S$ must be less than both $a$ and $b$, $S$ must equal $\boxed{\left( \sqrt{a}-\sqrt{b}\right)^2}$.
From the value of $S$, $x=\sqrt{2ab}-b$ and $y=\sqrt{2ab}-a$. Let the distance from the tangency point to $AD$ be $p$ and the distance from the tangency point to $AB$ be $q$. Then $p=r_1+x=r_1+\sqrt{2ab}-b$ and $q=r_1+y=r_1+\sqrt{2ab}-a$. It follows that $q=p+(b-a)$.
The locus must be a line segment that is included in the equation above. To find the full locus, we must find the endpoints. Note that $p$ and $q$ are both maximized when $r_1$ is maximized. Let $m=\min (a,b)$. Then $\max r_1=\frac m2$ so $\max p=\frac m2 +\sqrt{2ab}-b$ and $\max q=\frac m2 +\sqrt{2ab}-a$. Similarly, $p$ and $q$ are minimized when $r_2$ is maximized, which is when $r_1=S-\frac m2$.
Since any point between the endpoints is in the locus, the locus consists of the line segment connecting $\left( a-\frac m2, b-\frac m2 \right)$ and $\left( \frac m2 +\sqrt{2ab}-b,\frac m2 +\sqrt{2ab}-a\right)$.
Let the radius of the lower left circle be $r_1$, that of the upper circle be $r_2$, the horizontal component of the distance between the centers be $x$, and the vertical component of the distance between the centers be $y$. Also, letting $S=r_1+r_2$, we have the equations
\begin{eqnarray*} x&=&a-S\\ y&=&b-S\\ x^2+y^2&=&S^2 \end{eqnarray*}Substituting the values from the first and second equations into the third, $(a-S)^2+(b-S)^2=S^2\Rightarrow S^2-2(a+b)S+(a^2+b^2)=0$. The quadratic formula gives $S=a+b\pm \sqrt{2ab}=\left( \sqrt{a}\pm \sqrt{b}\right)^2$, and since $S$ must be less than both $a$ and $b$, $S$ must equal $\boxed{\left( \sqrt{a}-\sqrt{b}\right)^2}$.
From the value of $S$, $x=\sqrt{2ab}-b$ and $y=\sqrt{2ab}-a$. Let the distance from the tangency point to $AD$ be $p$ and the distance from the tangency point to $AB$ be $q$. Then $p=r_1+x=r_1+\sqrt{2ab}-b$ and $q=r_1+y=r_1+\sqrt{2ab}-a$. It follows that $q=p+(b-a)$.
The locus must be a line segment that is included in the equation above. To find the full locus, we must find the endpoints. Note that $p$ and $q$ are both maximized when $r_1$ is maximized. Let $m=\min (a,b)$. Then $\max r_1=\frac m2$ so $\max p=\frac m2 +\sqrt{2ab}-b$ and $\max q=\frac m2 +\sqrt{2ab}-a$. Similarly, $p$ and $q$ are minimized when $r_2$ is maximized, which is when $r_1=S-\frac m2$.
Since any point between the endpoints is in the locus, the locus consists of the line segment connecting $\left( a-\frac m2, b-\frac m2 \right)$ and $\left( \frac m2 +\sqrt{2ab}-b,\frac m2 +\sqrt{2ab}-a\right)$.
Hello. In the part 2 of the solution, instead of
$p=r_1+x$
$q=r_1+y$,
I think that the correct is
$p=\frac{r_1x}{S}+r_1$
$q=\frac{r_1y}{S}+r_1$.
This is because $x$ and $y$ are the components of the distance between the centers instead of the components of $r_1$.
Thanks
