# Maths: circles touching the x or y-axis

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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?

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I may be giving too much away, but remember that, when we factor $x^2+2gx+c=0$, we'll arrive at $(x+a)(x+b)=0$, where the following must be true: $a+b=2g$ and $ab=c$.

Edited by John
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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?

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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 $x$ must have a repeated root.

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