amy Posted March 12, 2013 Share Posted March 12, 2013 if x^{2}+y^{2}+2gx + 2fy +c =0 is on the x-axis, prove that g^{2} = c. I know that on the x-axis, y=0. x^{2} + (0)^{2} +2gx + 2f(0) +c =0 x^{2} + 2gx +c= 0 (which is a quadratic) but now i'm stuck. solving the quadratic with the quadratic equation formula doesn't help, so what should i do? Link to comment Share on other sites More sharing options...

John Posted March 12, 2013 Share Posted March 12, 2013 (edited) I may be giving too much away, but remember that, when we factor [math]x^2+2gx+c=0[/math], we'll arrive at [math](x+a)(x+b)=0[/math], where the following must be true: [math]a+b=2g[/math] and [math]ab=c[/math]. Edited March 13, 2013 by John Link to comment Share on other sites More sharing options...

amy Posted March 12, 2013 Author Share Posted March 12, 2013 well if i factorise by completing the square then i get x^{2} + 2gx + g^{2} =0 which gives me (x+g)(x+g) = 0 like as you said (x+a)(x+b) = 0 and g + g =2g (a + b = 2g as you said ) and ab = (g)(g) = g^{2 }=c , but why is it equal to c? Link to comment Share on other sites More sharing options...

Nehushtan Posted March 13, 2013 Share Posted March 13, 2013 well if i factorise by completing the square then i get x^{2} + 2gx + g^{2} =0 which gives me (x+g)(x+g) = 0 like as you said (x+a)(x+b) = 0 and g + g =2g (a + b = 2g as you said ) and ab = (g)(g) = g^{2 }=c , but why is it equal to c? Here’s another hint: If the circle is tangent to the x-axis, the quadratic equation in [latex]x[/latex] must have a repeated root. Link to comment Share on other sites More sharing options...

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