Four Kissing Circles
What is the radius of the small, red circle?
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3 solutions
hello, can please write the solution using the decartes theorem
By Descarte's Circle Theorem,
To get the curvature of the three circles, just get the reciprocal of their radii. Those shall be your k 1, k 2, and k_3.
k_4=(1/2)+(1/2)+(1/3)+2sqrt[(1/2 * 1/2) + (1/2 * 1/3) + (1/2 * 1/3)]
k_4=4/3 + 2sqrt[(1/4) + (1/6) + (1/6)]
k_4=4/3 + 2sqrt(7/12)
k_4=4/3 + sqrt(21)/3
k_4=[4+sqrt(21)]/3
Note that since the circle is externally tangent to the three larger circles, its curvature is positive.
[4+sqrt(21)]/3=1/r
r=3/[4+sqrt(21)]
Rationalize the denominator,
r={3[4-sqrt(21)]}/(-5)
r=[12-3sqrt(21)]/(-5)
r=[3sqrt(21)-12]/5
r={3[sqrt(21)-4]}/5
how did you get from 4/3 + 2sqrt(7/12) to 4/3 +sqrt(21)/3 ?
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To get rid of the fraction under the radical, multiply 7/12 by 3/3. This results in a perfect square in the denominator that can be removed from the radical.
4/3 + 2√(7/12) = 4/3 + 2√(21/36) = 4/3 + (2√21)/6 = 4/3 +3√21
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Sorry, messed up the last term:
4/3 + 2√(7/12) = 4/3 + 2√(21/36) = 4/3 + (2√21)/6 = 4/3 +√(21)/3
A direct application of Descartes' Circle Theorem
r 4 1 r 4 1 r 4 r 4 = r 1 1 + r 2 1 + r 3 1 ± 2 r 1 r 2 1 + r 2 r 3 1 + r 3 r 1 1 = 2 1 + 2 1 + 3 1 ± 2 ( 2 ) ( 2 ) 1 + ( 2 ) ( 3 ) 1 + ( 3 ) ( 2 ) 1 = 3 4 + 2 1 2 7 = 3 4 + 3 7 = 3 4 + 2 1 = 2 1 + 4 3 = 5 3 ( 2 1 − 4 )