Cartesian coordinates on 2 circles

Question

What are the coordinates of points $j$ and $k$ in terms of $a$, $b$, and $c$?

The curves are circular arcs with the center of the circles at $y = 0$.

Answer 1

(copying and editing John's answer, since the edit to that answer was rejected)

The line from $j$ through $k$ passes through the circle center which is on the $x$-axis; let's say it's at $(R+a, 0)$, so that $R$ is the radius of the inner circle.

The angle from the negative $x$ axis clockwise to the line $jk$, the acute angle in this picture...let's call that $t$ (radians). Then $$ Rt = c \\ (R+a)t = b $$ Hence $$ c + at = b, $$ and we get $$ t = \frac{b-c}{a}. $$ We also see that $$ R = \frac{c}{t} = \frac{ca}{b-c}. $$

The location of point $k$ is then $$ k = (R+a, 0) + R(-\cos t, \sin t) $$ and for point $j$, we have $$ j = (R+a, 0) + (R+a) (-\cos t, \sin t). $$

If we want these in terms of only $a$, $b$, and $c$ then $$ k = (\frac{ca}{b-c}+a, 0) + \frac{ca}{b-c}(-\cos \frac{b-c}{a}, \sin \frac{b-c}{a}) \\ j = (\frac{ca}{b-c}+a, 0) + (\frac{ca}{b-c}+a) (-\cos \frac{b-c}{a}, \sin \frac{b-c}{a}) $$

Answer 2

The line from $i$ through $j$ passes through the circle center which appears, from the drawing, to be on the $x$-axis; let's say it's at $(R+a, 0)$, so that $r$ is the radius of the inner circle.

The angle from the negative $x$ axis clockwise to the line $ij$, the acute angle in this picture...let's call that $t$. Then $$ Rt = c \\ (R+a)t = b $$ Hence $$ c + at = d, $$ and we get $$ t = \frac{b-c}{a}. $$ We also see that $$ R = \frac{c}{t} = \frac{ca}{b-c}. $$

The location of point $k$ is then $$ k = (R+a, 0) + R(-\cos t, \sin t) $$ and for point $j$, we have $$ j = (R+a, 0) + (R+a) (-\cos t, \sin t). $$