document.write( "Question 258000: What is the points of intersection of these equations
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document.write( "x-2y-8=0
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document.write( "x^2=y^2-12x+6y+29=0 \n" );
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Algebra.Com's Answer #189894 by Greenfinch(383)![]() ![]() You can put this solution on YOUR website! Second equation is a circle, so reorganise it \n" ); document.write( "(x^2 + 12 x) - (y^2 - 6y)+ 29 = 0 and then complete the square \n" ); document.write( "(x^2 + 12x + 36) - ( y^2 - 6y + 9) - 45 + 29 = 0 which becomes \n" ); document.write( "(x + 6)^2 - (y - 3)^2 = 4^2 which is a circle centre at -6, 3 and radius 4 \n" ); document.write( "other equation is y = (1/2)x - 4 or x = 2y + 8 \n" ); document.write( "Probably solve for y is easier \n" ); document.write( "(2y + 14)^2 - (y - 3)^2 = 4^2 \n" ); document.write( "4y^2 + 56 y + 196 -(y^2 - 6y +9) = 16 \n" ); document.write( "3y^2 + 62y + 187 = 16 \n" ); document.write( "3y^2 + 62y + 171 = 0
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