10Asupp1

# 10Asupp1 - MATH 10A Section 3.7 34 Problem Find the points...

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Unformatted text preview: MATH 10A Section 3.7 # 34 Problem Find the points where the circles x2 + y 2 = 1 and (x − 3)2 + y 2 = 4 share a common tangent line. . . . . . . . (3 − p, q ) ............................ . ........................ . .... . .... .. .. . ...... . . .... . ... ...... .. ...... ... . .. . ... ... ........ ........ . . . .. ... ..... ..... ... . . .. ..... ..... .. ... . . .. ... . .. ... ..... .. . . .. ..... .. ... . . . .. .. . . .. ... ..... . ......... . .. .. .. . . .. ..... ... . .. .. ....... .. .. .. .. .. .. .. ... .. . .. 2 (−a, b)..................... .. ..... . ........ . .. . .. . ... . . ... . . .. . . ... . . ... . .. . . .. ..... .. . . . .. ... ... . . . ... . .. . . . . . q. ..... ... . . ..... .... ... . . .. . . .. ... . .. .. . . . . . . .. . . . . ..... . . .1. . . ..... .. . . . .. ... .. . . .. . . . ... . . b . .. . . . ..... .. . . ..... . . .. . .. ... .. . . . . .. . . .. . . . .. . ..... . . ... . ..... . .. . .. . . . ... . .. . . . . . . . ... . . . . . ..... . .. . ..... ... . .. . . .... . ..................................................................................................................................................................................................... ............................................................................................................................. ....................................................................... . . . . . . . . . . . . . . . . . . . . p .. . . a. . .. . . . .. . . . . . .. . .. . .. .. . .. . .. . . .. . . .. . .. . . .. . . . .. .. ... . ... . . . . . ... ... . ... . . ... .. . .. . ..... . ....... ..... . . ...... . . .. .......... ........ .. .. .. . . . .. .. .. . . .. ... . .. ... . .. . .. ... . ... . ... ... . ... ... . ... . ... . ... . ... . . ... .... ... .... . . ..... .... ..... . .... . ......... ............. . ............. ............ . ..... . • • • • Solution: Using similar triangles, we get b q = 1 2 and a p =, 1 2 hence p = 2a and q = 2b. So the point on the larger circle has coordinates (3 − 2a, 2b). The slope of the line through the points (−a, b) and (3 − 2a, 2b) is the slope of the tangent line to the small circle at (−a, b). By implicit diﬀerentiation of the smaller circle, 2x + 2y dy = 0, dx so dy x =− . dx y Thus the slope of the tangent line at the point (−a, b) is a . Therefore, setting b a 2b − b =, (3 − 2a) − (−a) b or b2 = 3a − a2 . Since (−a, b) is a point on the unit circle, then a2 + b2 = 1. Using the substitution √ √ 1 b2 = 1 − a2 , we get 1 − a2 = 3a − a2 , so a = 3 , and b = 1 − a2 = 1 − 1 = 38 . 9 1 Therefore, (−a, b) = (− 3 , √ 8 ), 3 implies b a =, 3−a b and (3 − p, q ) = (3 − 2a, 2b) = ( 7 , 2 3 8 ). 3 √ ...
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