if x²+y²+2gx + 2fy +c =0 is on the x-axis, prove that g² = c.

 

 

I know that on the x-axis, y=0.

x² + (0)² +2gx + 2f(0) +c =0

x² + 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?

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, 201312 yr by John

well if i factorise by completing the square then i get

x² + 2gx + g² =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) = g² =c , but why is it equal to c?

well if i factorise by completing the square then i get

x² + 2gx + g² =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) = g² =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.