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Suppose that there are four points $A, B, C, D$. A circle of radius $r_A$ surrounds point $A$, a circle of radius $r_C$ surrounds point $C$, and a circle of radius $|DB|$ surrounds point $D$. $AB||BC$, $|AB|=|BC|$. Circle $D$ intersects circles $A$ and $C$ exactly once per circle, at different points, in addition to intersecting with point $B$.

graphical illustration

How to solve $|DB|$ algebraically (preferably without iterating) knowing only the values of $r_A$, $r_C$ and $|AB|$?

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2 Answers 2

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Your constraints (first 2 are law of cosines):

$$|DA|^2 = |DB|^2 + |AB|^2 - 2|AB||DB|\cos(x)$$ $$|DC|^2 = |DB|^2 + |BC|^2 - 2|BC||DB|\cos(y)$$

$$x + y = \pi$$ $$|DA| = r_A + |DB|$$ $$|DC| = r_C + |DB|$$ $$|AB| = |BC|$$

Substitute the last 4 into the first 2:

$$(r_A + |DB|)^2 = |DB|^2 + |AB|^2 - 2|AB||DB|\cos(x)$$ $$(r_C + |DB|)^2 = |DB|^2 + |AB|^2 - 2|AB||DB|\cos(\pi - x)$$

Now you have 2 equations and 2 variables, x and |DB|. I suggest writing $\cos(\pi - x) = \cos(\pi)\cos(x) + \sin(\pi)\sin(x) = -\cos(x)$.

$$(r_A + |DB|)^2 = |DB|^2 + |AB|^2 - 2|AB||DB|\cos(x)$$ $$(r_C + |DB|)^2 = |DB|^2 + |AB|^2 + 2|AB||DB|\cos(x)$$

Then just add and solve the resulting equation. The quadratic terms look like they'll cancel out so it should be straightfoward.

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Hint: Let $p_1$ be the POI (point of intersection) of circle $A$ with circle $D$. Let $p_2$ be the POI of circle $C$ with circle $D$.

You can easily find $|p_1 B|$ and $|p_2 B|$; also $\angle ABp_1$ and $\angle CBp_2$.

Then you have an isosceles triangle $\triangle DBp_1$ and you know $|p_1 B|$. What can you do now?

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