Externally Tangent Circles

Question

Given two distinct points $A, B$, what is the set of possible tangency points between externally tangent circles $o_1; o_2$ such that $o_1$ is tangent to the line $AB$ at $A$, and $o_2$ is tangent to the line $AB$ at $B$.

Any hints on how to do this?

I think the picture look like this

Thanks

Answer 1

Good picture.

Without loss of generality, let point $A$ be the origin $(0,0)$ and let point $B$ be $(b,0)$.

They are joined by vector $\vec {AB} = \left (\matrix{b \\ 0} \right)$.

Let $\vec {n}=\left (\matrix{0\\ 1} \right)$ be the unit vector perpendicular to $\vec {AB}$.

Then $\vec {OO_1}= \lambda \vec n = \left (\matrix{0\\ \lambda} \right)$ and $\vec {OO_2}= \vec {OB} + \mu \vec n=\left (\matrix{b \\ 0} \right)+\left (\matrix{0\\ \mu} \right) = \left (\matrix{b\\ \mu} \right)$.

But $\lambda$ and $\mu$ will be the radii of the circles and the distance $|O_1O_2|=\lambda + \mu$

$\vec {O_1O_2}= \vec{OO_2}-\vec {OO_1}= \left (\matrix{b \\ \mu-\lambda} \right)$

$b^2+(\lambda-\mu)^2=(\lambda+\mu)^2$

$b^2+\lambda^2-2\lambda \mu +\mu^2=\lambda^2+2\lambda \mu +\mu^2$

$b^2 = 4\lambda \mu$

Your point of tangency divides $O_1O_2$ in the ratio $\lambda:\mu$,

so it has position vector $\vec {OP}=\vec {OO_1} + \frac {\lambda}{\lambda+\mu} \vec {O_1O_2}$

$\vec {OP}=\left (\matrix{0 \\ \lambda} \right)+\frac {\lambda}{\lambda+\mu}\left (\matrix{b \\ \mu-\lambda} \right)$

$\vec {OP}=\frac {1}{\lambda+\mu} \left (\matrix{0 \\ \lambda (\lambda +\mu)} \right)+\frac {1}{\lambda+\mu} \left (\matrix{\lambda b \\ \lambda \mu-\lambda^2} \right)$

$\vec {OP}=\frac {1}{\lambda+\mu} \left (\matrix{0 \\ \lambda ^2+\lambda \mu)} \right)+\frac {1}{\lambda+\mu} \left (\matrix{\lambda b \\ \lambda \mu-\lambda^2} \right)$

$\vec {OP}=\frac {1}{\lambda+\mu} \left (\matrix{\lambda b \\ 2\lambda \mu} \right)$

Answer 2

Hints: Start by showing that the line from the midpoint of $AB$ to the point of tangency is perpendicular to $O_1 O_2$.

Then argue that the distance from the point of tangency to $AB$ is $\frac{1}{2} |AB|$, and hence the locus is a circle centered at the midpoint of $AB$ with radius $\frac{1}{2}|AB|$.

Illustrated here: