document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1151274>Question 1151274</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose a circle of radius 1 is internally tangent to two circles of radius 2. What is the blue shaded area?<BR>\n" );
document.write( "(please refer to the link below as it contains the image for the question)<BR>\n" );
document.write( "https://i.imgur.com/l66o6SR.png </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #772971 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=772971\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Label the three circles A, B, C<BR>\n" );
document.write( "Circle A is the circle with radius 1<BR>\n" );
document.write( "Circles B and C are the circles with radius 2.<BR>\n" );
document.write( "Circle A is internally tangent to circles B and C.\r<BR>\n" );
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document.write( "Place the center of circle A at the origin (0,0), which I'll call point P. <BR>\n" );
document.write( "The equation of this circle is x^2+y^2 = 1\r<BR>\n" );
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document.write( "Mark two new points Q and R, such that,<BR>\n" );
document.write( "Q = (-1,0)<BR>\n" );
document.write( "R = (1,0)<BR>\n" );
document.write( "These points are on the west and east sides of circle A, which will be the centers of circles B and C. \r<BR>\n" );
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document.write( "The equation of circle B is<BR>\n" );
document.write( "(x+1)^2 + y^2 = 4<BR>\n" );
document.write( "The equation of circle C<BR>\n" );
document.write( "(x-1)^2 + y^2 = 4\r<BR>\n" );
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document.write( "Solve the system<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=system%28%28x%2B1%29%5E2+%2B+y%5E2+=+4%2C%28x-1%29%5E2+%2B+y%5E2+=+4%29 ALIGN=MIDDLE  ALT=\"system%28%28x%2B1%29%5E2+%2B+y%5E2+=+4%2C%28x-1%29%5E2+%2B+y%5E2+=+4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "to end up with the two points of intersection S and T<BR>\n" );
document.write( "S = (0, sqrt(3))<BR>\n" );
document.write( "T = (0, -sqrt(3))<BR>\n" );
document.write( "I'll let you fill in the details for this part as an exercise. If you're stuck on how to do this, then please let me know. A hint is that you can subtract the equations and have the y^2 terms cancel out, leaving you with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B1%29%5E2-%28x-1%29%5E2+=+0 ALIGN=MIDDLE  ALT=\"%28x%2B1%29%5E2-%28x-1%29%5E2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >. \r<BR>\n" );
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document.write( "So far we have this diagram<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/qvDIy8U.png\"><BR>\n" );
document.write( "P = (0,0)<BR>\n" );
document.write( "Q = (-1,0)<BR>\n" );
document.write( "R = (1,0)<BR>\n" );
document.write( "S = (0, sqrt(3))<BR>\n" );
document.write( "T = (0, -sqrt(3))\r<BR>\n" );
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document.write( "Draw in triangle STR and then pull it away from the diagram. Relabel the vertices to A,B,C temporarily<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/fu3VQSA.png\"><BR>\n" );
document.write( "the lowercase letters a,b,c are opposite their respective uppercase counterparts A,B,C.\r<BR>\n" );
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document.write( "Let's find angle C. Use the law of cosines.<BR>\n" );
document.write( "c^2 = a^2 + b^2 - 2*a*b*cos(C)<BR>\n" );
document.write( "(2sqrt(3))^2 = 2^2 + 2^2 - 2*2*2*cos(C)<BR>\n" );
document.write( "12 = 8 - 8*cos(C)<BR>\n" );
document.write( "12-8 = -8*cos(C)<BR>\n" );
document.write( "4 = -8*cos(C)<BR>\n" );
document.write( "cos(C) = -1/2<BR>\n" );
document.write( "C = arcos(-1/2)<BR>\n" );
document.write( "C = 120 <BR>\n" );
document.write( "This means angle SRT is 120 degrees. If we just focus on triangle SRT, then we can refer to this as \"angle R\" in shorthand.\r<BR>\n" );
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document.write( "Now compute the area of the blue shaded region below<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/3V6VTS4.png\"><BR>\n" );
document.write( "which represents the circular sector TQS (centered at point R).\r<BR>\n" );
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document.write( "area of sector = (angle/360)*pi*r^2<BR>\n" );
document.write( "area of sector TQS = (angle SRT/360)*pi*(QR)^2<BR>\n" );
document.write( "area of sector TQS = (120/360)*pi*2^2<BR>\n" );
document.write( "area of sector TQS = (1/3)*pi*4<BR>\n" );
document.write( "area of sector TQS = (4/3)*pi<BR>\n" );
document.write( "We will use this value later, so let M = (4/3)*pi\r<BR>\n" );
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document.write( "Compute the area of triangle SRT using the side-angle-side (SAS) triangle area rule<BR>\n" );
document.write( "area of triangle = (1/2)*(side1)*(side2)*sin(included angle)<BR>\n" );
document.write( "area of triangle SRT = (1/2)*(SR)*(RT)*sin(R)<BR>\n" );
document.write( "area of triangle SRT = (1/2)*(2)*(2)*sin(120)<BR>\n" );
document.write( "area of triangle SRT = 2*sin(120)<BR>\n" );
document.write( "area of triangle SRT = 2*sqrt(3)/2<BR>\n" );
document.write( "area of triangle SRT = sqrt(3)<BR>\n" );
document.write( "We will use this value later, so let N = sqrt(3)\r<BR>\n" );
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document.write( "We now have this diagram<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/PerunV7.png\"><BR>\n" );
document.write( "I have added the green triangle over the blue circular sector\r<BR>\n" );
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document.write( "If we take away the triangle through subtraction, then we end up with this region here<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/F4Gbbo6.png\"><BR>\n" );
document.write( "that area is exactly M-N =  (4/3)*pi -  sqrt(3)<BR>\n" );
document.write( "as its the area of the circular sector minus the area of the triangle.\r<BR>\n" );
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document.write( "Double this result,<BR>\n" );
document.write( "2*(M-N) = 2*((4/3)*pi -  sqrt(3))<BR>\n" );
document.write( "2*(M-N) = (8/3)*pi -  2sqrt(3)<BR>\n" );
document.write( "this expression represents the exact area of this blue region shown below<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/ne9r0aN.png\"><BR>\n" );
document.write( "The doubling can be done due to the symmetry along the y axis.\r<BR>\n" );
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document.write( "Finally, the last step is to subtract off the area of the inner circle (circle A) that has radius 1. The area of this small circle is pi*r^2 = pi*1^2 = pi square units.\r<BR>\n" );
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document.write( "So we have,<BR>\n" );
document.write( "2*(M-N) - (area of circle A) = (8/3)*pi -  2sqrt(3) - pi<BR>\n" );
document.write( "2*(M-N) - (area of circle A) = (8/3)*pi -  2sqrt(3) - 3pi/3<BR>\n" );
document.write( "2*(M-N) - (area of circle A) = <font color=red size=4>(5/3)*pi -  2sqrt(3)</font>\r<BR>\n" );
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document.write( "This represents the final shaded blue region we are after<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/hH49f0d.png\">\r<BR>\n" );
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document.write( "Therefore, the <font color=red size=4>final answer is choice B</font>\r<BR>\n" );
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