document.write( "Question 1202363: A unit circle has its center at (5,0) and a second circle with a radius of 2 units has its center at (11,0) as shown. A common internal tangent to the circles intersects the x-axis at Q(a,0). What is the value of a? \n" ); document.write( "

Algebra.Com's Answer #837124 by ikleyn(52873) $\"\"$ $\"About$

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\n" ); document.write( "A unit circle has its center at (5,0) and a second circle with a radius of 2 units
\n" ); document.write( "has its center at (11,0) as shown. A common internal tangent to the circles intersects
\n" ); document.write( "the x-axis at Q(a,0). What is the value of a?
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document.write( "Let A = (5,0) and E = (11,0)  be the centers of the circles;\r\n" );
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document.write( "let B and D be the tangent points on greater and smaller circles, respectively;\r\n" );
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document.write( "Q = (a,0) is the intersection point of the common internal tangent to the circles\r\n" );
document.write( "with the x-axis.\r\n" );
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document.write( "Draw the radii AB and ED to tangent points and consider triangles ABQ and QDE.\r\n" );
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document.write( "These triangles are right-angled triangles with right angles at the tangent points B and D.\r\n" );
document.write( "They have congruent acute angles at the intersection point Q.\r\n" );
document.write( "Hence, these triangles are similar.\r\n" );
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document.write( "Their corresponding sides are the radii AB = 1 and ED = 2 (the legs).\r\n" );
document.write( "Their other corresponding sides are AQ and EQ (the hypotenuses),\r\n" );
document.write( "so we can write \r\n" );
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document.write( "     =  = 2.\r\n" );
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document.write( "From the other side,  AQ + EQ = AE = 11 - 5 = 6 units.\r\n" );
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document.write( "So, the intersection point Q divides the segment AE of the length 6 units \r\n" );
document.write( "in segments  EQ  and  AQ  in proportion  2:1.\r\n" );
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document.write( "It implies that  AQ  is one third of  AE=6;  in other words,  AQ = 2 units.\r\n" );
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document.write( "Hence, the value \"a\" under the problem's question is  a = 5 + 2 = 7 units.    ANSWER\r\n" );
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