document.write( "Question 1067878: Find the equation of a circle that has radius length {sqrt(30)} and is tangent to the line 3x+y-5=0 at the point (-1,8). \n" ); document.write( "

Algebra.Com's Answer #683110 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "Find the equation of a circle that has radius length {sqrt(30)} and is tangent to the line 3x+y-5=0 at the point (-1,8).
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\n" ); document.write( "\n" ); document.write( "I think that more short (and more straightforward) solution is possible.\r
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document.write( "The idea is to draw (to write the equation of) the straight line perpendicular to the given line through the given point (-1,8), \r\n" );
document.write( "and then to take its intersection with the circle of the radius  centered at the given point (-1,8).\r\n" );
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document.write( "Plot y = 5-3x (the given line, red), y =  (the perpendicular at (-1,8), green)\r\n" );
document.write( "and the circle centers (intersection points of the green straight line with the arcs, blue and purple arcs)\r\n" );
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document.write( "1.  The straight line perpendicular to the given line  3x+y-5=0  at the point (-1,8) is\r\n" );
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document.write( "    -(x-(-1)) + 3(y-8) = 0,   or  (which is the same)  -x - 1 + 3y - 24 = 0  or  (which is equivalent)  x = 3y-25.\r\n" );
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document.write( "2.  The circle of the radius  centered at (-1,8) is\r\n" );
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document.write( "     = .\r\n" );
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document.write( "3.  The centers of two circles we are searching for, are the intersection points, i.e. the solutions of the system\r\n" );
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document.write( "     x = 3y-25,            (1)\r\n" );
document.write( "      = .    (2)\r\n" );
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document.write( "     To solve the system, substitute (1) into (2), replacing x. You will get\r\n" );
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document.write( "      = 30,\r\n" );
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document.write( "     9y^2 - 144y + 576 + y^2 - 16y + 64 = 30,\r\n" );
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document.write( "     10y^2 - 160y + 610 = 0,\r\n" );
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document.write( "      =  =  = .\r\n" );
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document.write( "     Thus the centers are\r\n" );
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document.write( "     a)   =  ,   = 3y-25 = ,   and\r\n" );
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document.write( "     b)   =  ,   = 3y-25 = .\r\n" );
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\n" ); document.write( "\n" ); document.write( "Having the centers and the radius r = $\"sqrt%2830%29\"$ , everybody can write the standard equations for the two circles.\r
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