Se o raio do círculo de centro B é 2, então o raio do outro círculo é:
A)6
B)5
C)4,5
D)5,5
E)6,5
Anexos:
![](https://pt-static.z-dn.net/files/da3/8c0a645911cd935d3e39615450609df6.jpg)
Soluções para a tarefa
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๏ Formula do quadrado da soma:
![\mathsf{\left(a+b\right)^2=a^2+2ab+b^2} \mathsf{\left(a+b\right)^2=a^2+2ab+b^2}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%28a%2Bb%5Cright%29%5E2%3Da%5E2%2B2ab%2Bb%5E2%7D)
๏ Formula do quadrado da diferença:
![\mathsf{\left(a-b\right)^2=a^2-2ab+b^2} \mathsf{\left(a-b\right)^2=a^2-2ab+b^2}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cleft%28a-b%5Cright%29%5E2%3Da%5E2-2ab%2Bb%5E2%7D)
= = = = =
Dada a relação na Figura em Anexo, formando um triangulo retângulo, e sabendo que o raio do círculo de centro B vale 2 u.m. Aplicando o teorema de Pitágoras:
![\mathsf{H^2=CA^2+CB^2}\\\\\mathsf{\left(R_A+R_B\right)^2=\left(R_A-R_B\right)^2+6^2}\\\\\mathsf{\left(R_A+2\right)^2=\left(R_A-2\right)^2+6^2}\\\\\mathsf{R_A^2+\left(2\cdot 2\cdot R_A\right)+2^2=R_A^2-\left(2\cdot 2\cdot R_A\right)+2^2+6^2}\\\\\mathsf{\diagup\!\!\!\!\!R_A^2+4R_A+\diagup\!\!\!\!4=\diagup\!\!\!\!\!R_A^2-4R_A+\diagup\!\!\!\!4+36}\\\\\mathsf{4R_A=-4R_A+36}\\\\\mathsf{4R_A+4R_A=36}\\\\\mathsf{8R_A=36}\\\\\mathsf{R_A=\dfrac{36}{8}}\\\\\boxed{\mathsf{R_A=4,5}}\: \: \checkmark \mathsf{H^2=CA^2+CB^2}\\\\\mathsf{\left(R_A+R_B\right)^2=\left(R_A-R_B\right)^2+6^2}\\\\\mathsf{\left(R_A+2\right)^2=\left(R_A-2\right)^2+6^2}\\\\\mathsf{R_A^2+\left(2\cdot 2\cdot R_A\right)+2^2=R_A^2-\left(2\cdot 2\cdot R_A\right)+2^2+6^2}\\\\\mathsf{\diagup\!\!\!\!\!R_A^2+4R_A+\diagup\!\!\!\!4=\diagup\!\!\!\!\!R_A^2-4R_A+\diagup\!\!\!\!4+36}\\\\\mathsf{4R_A=-4R_A+36}\\\\\mathsf{4R_A+4R_A=36}\\\\\mathsf{8R_A=36}\\\\\mathsf{R_A=\dfrac{36}{8}}\\\\\boxed{\mathsf{R_A=4,5}}\: \: \checkmark](https://tex.z-dn.net/?f=%5Cmathsf%7BH%5E2%3DCA%5E2%2BCB%5E2%7D%5C%5C%5C%5C%5Cmathsf%7B%5Cleft%28R_A%2BR_B%5Cright%29%5E2%3D%5Cleft%28R_A-R_B%5Cright%29%5E2%2B6%5E2%7D%5C%5C%5C%5C%5Cmathsf%7B%5Cleft%28R_A%2B2%5Cright%29%5E2%3D%5Cleft%28R_A-2%5Cright%29%5E2%2B6%5E2%7D%5C%5C%5C%5C%5Cmathsf%7BR_A%5E2%2B%5Cleft%282%5Ccdot+2%5Ccdot+R_A%5Cright%29%2B2%5E2%3DR_A%5E2-%5Cleft%282%5Ccdot+2%5Ccdot+R_A%5Cright%29%2B2%5E2%2B6%5E2%7D%5C%5C%5C%5C%5Cmathsf%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21R_A%5E2%2B4R_A%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%214%3D%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21R_A%5E2-4R_A%2B%5Cdiagup%5C%21%5C%21%5C%21%5C%214%2B36%7D%5C%5C%5C%5C%5Cmathsf%7B4R_A%3D-4R_A%2B36%7D%5C%5C%5C%5C%5Cmathsf%7B4R_A%2B4R_A%3D36%7D%5C%5C%5C%5C%5Cmathsf%7B8R_A%3D36%7D%5C%5C%5C%5C%5Cmathsf%7BR_A%3D%5Cdfrac%7B36%7D%7B8%7D%7D%5C%5C%5C%5C%5Cboxed%7B%5Cmathsf%7BR_A%3D4%2C5%7D%7D%5C%3A+%5C%3A+%5Ccheckmark)
= = = = =
Resposta: Letra C
๏ Formula do quadrado da diferença:
= = = = =
Dada a relação na Figura em Anexo, formando um triangulo retângulo, e sabendo que o raio do círculo de centro B vale 2 u.m. Aplicando o teorema de Pitágoras:
= = = = =
Resposta: Letra C
Anexos:
![](https://pt-static.z-dn.net/files/d17/5e31a117ae9d373ad38531a91114a34a.jpg)
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