document.write( "Question 1143657: Find the equation of the circle that passes through the points of intersection of the circles x²+y²=2x, x²+y²=2y and has its center on the line y=2. \n" ); document.write( "

Algebra.Com's Answer #764526 by ikleyn(52802) $\"\"$ $\"About$

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document.write( "(1)  Since the left sides are identical in given equations of the circles, their right sides are equal for intersection points\r\n" );
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document.write( "         x = y.\r\n" );
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document.write( "(2)  Substituting  x=y  into the circles equations gives\r\n" );
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document.write( "         x^2 + x^2 = 2x  ====>  2x^2 = 2x  ====>  2x^2 - 2x = 0  ====>  2x(x-1) = 0.\r\n" );
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document.write( "     The roots are  x= 1  and  x= 0,  so the intersection points are  A=(1,1)  and  B=(0,0).\r\n" );
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document.write( "(3)  The center of the third circle lies on the perpendicular bisector to the segment AB.\r\n" );
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document.write( "     An equation of this perpendicular bisector is  y-0.5 = -(x-0.5) = -x + 0.5.\r\n" );
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document.write( "(4)  At the same time, the center of the third circle lies on the line  y= 2.\r\n" );
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document.write( "     So, the center of the third circle is at\r\n" );
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document.write( "         2 - 0.5 = -x + 0.5,\r\n" );
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document.write( "     which implies  x= -1.\r\n" );
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document.write( "     Thus  the center of the third circle is the point  P = (-1,2).\r\n" );
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document.write( "     The radius of the third circle is the distance of the point P from the origin (0,0), i.e.   =  = .\r\n" );
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document.write( "(5)  Thus the equation of the circle under the question is\r\n" );
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document.write( "          +  = ,\r\n" );
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document.write( "     or, equivalently,\r\n" );
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document.write( "          +  = 5.    ANSWER\r\n" );
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