This is actually a CAD design I'm working on. Points $A$, $B$, and $C$ are fixed. Need $R_2$ when circles are externally tangent.
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use Pythagoras Thm and calculate https://www.wolframalpha.com/input?i=solve+%289.463%2B28.465%29%5E2%2B%28R%2B7.108%29%5E2%3D%2828.465%2BR%29%5E2 . – janmarqz Dec 27 '22 at 18:35
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We have a right triangle with hypotenuse $R_1 + R_2$ and legs $R_1 + 9.463$ and $7.108 + R_2$. By the Pythagorean theorem, $$ (R_1 + R_2)^2 = (R_1 + 9.463)^2 + (7.108 + R_2)^2. $$ When you plug in $28.465$ for $R_1$, you are left with an equation in $R_2$ that an online calculator could solve.
Sam Freedman
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4In fact, the quadratic term is the same on both sides, so this is actually a linear equation! – diracdeltafunk Dec 27 '22 at 18:34
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The vertical leg of the right triangle has length $9.463 + R_1 = 9.463 + 28.465 = 37.928$. The horizontal leg of the right triangle has length $7.108 + R_2$. The hypotenuese of the right triangle has length $R_1 + R_2 = 28.465 + R_2$. So, by the Pythagorean theorem, we have
$$37.928^2 + (7.108 + R_2)^2 = (28.465 + R_2)^2$$
which simplifies to (approximately)
$$678.801 - 42.714 R_2 = 0.$$
Thus, $$R_2 \approx 15.892.$$
diracdeltafunk
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