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I have circle A, point B and circle C. Circle A and Circle C do not intersect. Given point B, can I find a circle, B, which is externally tangential to A, does not intersect circle C and has the point B on its circumference?

As far as I can tell, there will be a set of solutions to the right (in the drawing) and a set to the left? I am looking for the maximum radius of circle B where this is possible.

Circles A and C, point B and a possible solution for circle B.

enter image description here

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  • $\begingroup$ Have you tried using coordinate geometry? $\endgroup$ – Qwerty Oct 13 '16 at 20:13
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If there is a straight line passing through $B$ and between the two circles, then of course that line is the solution with highest (infinite) radius. If such line does not exist, then the highest radius circle must be tangent to both circles. This is a variation of Apollonius' problem and can be solved in many different ways.

Let then $O$ be the center of a circle through $B$, internally tangent to the circle of center $C$ and externally tangent to the circle of center $A$. If $r_A$ and $r_C$ are the radii of the circles, we then have $OB-OC=r_C$ and $OA-OB=r_A$. Thus point $O$ lies on the intersection of two hyperbolas: the first one having foci $B$, $C$ and major axis $r_C$, the second one having foci $B$, $A$ and major axis $r_A$ (see diagram below).

Of the four intersection points between the hyperbolas, only two satisfy the conditions of the problem and the one farther from $B$ is the center of the required circle (red in the diagram).

enter image description here

To find the coordinates $(x,y)$ of $O$ and the radius $r$ of the tangent circle one must then solve the following set of equations: $$ \begin{cases} (x-x_C)^2+(y-y_C)^2=(r-r_C)^2\\ (x-x_A)^2+(y-y_A)^2=(r+r_A)^2\\ (x-x_B)^2+(y-y_B)^2=r^2\\ \end{cases} $$

Though apparently of eighth degree, this system has only two solutions and can be solved quite easily. A symbolic expression for the solution is however too cumbersome to be reported here.

Another geometrical solution, which can be done with compass and straightedge, involves inversion by an arbitrary circle centered at $B$ (gray circle in the diagram below). The two given circles are transformed by inversion into two other circles (dashed), while any circle through $B$ is transformed by inversion into a line. But it is easy to construct the two lines tangent to the inverted circles and leaving them on opposite sides (dashed lines): by inverting those lines we get then two tangent circles passing through $B$ (red and blue circles).

enter image description here

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