RT and circles

Solve right triangle if the radius of inscribed circle is r=9 and radius of circumscribed circle is R=23.

Result

a =  37.83
b =  26.17
c =  46

Solution:

R=c2 c=2R=46  r=a+b−c2 a+b=64 a2+b2=2116  2a2−128a+1980=0  p=2;q=−128;r=1980 D=q2−4pr=1282−4⋅2⋅1980=544 D>0  a1,2=−q±D2p=128±5444=128±4344 a1,2=32±5.83095189485 a1=37.8309518948 a2=26.1690481052   Factored form of the equation:  2(a−37.8309518948)(a−26.1690481052)=0 R = \dfrac{c}{2} \ \\ c = 2 R = 46 \ \\ \ \\ r = \dfrac{ a+b-c}{2} \ \\ a + b = 64 \ \\ a^2 + b^2 = 2116 \ \\ \ \\ 2a^2 -128a +1980 =0 \ \\ \ \\ p=2; q=-128; r=1980 \ \\ D = q^2 - 4pr = 128^2 - 4\cdot 2 \cdot 1980 = 544 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 128 \pm \sqrt{ 544 } }{ 4 } = \dfrac{ 128 \pm 4 \sqrt{ 34 } }{ 4 } \ \\ a_{1,2} = 32 \pm 5.83095189485 \ \\ a_{1} = 37.8309518948 \ \\ a_{2} = 26.1690481052 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (a -37.8309518948) (a -26.1690481052) = 0 \ \\
b=(−(−128)−(23.3238075794))/(2⋅ (2))=26.17b=(-(-128) - (23.3238075794))/ (2 \cdot \ (2))=26.17
c=2⋅ 23=46c=2 \cdot \ 23=46

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