[ASK] A Circle Which Touches the X-Axis at 1 Point

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Monoxdifly
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The circle $$x^2+y^2+px+8y+9=0$$ touches the X-axis at one point. The center of that circle is ...
a. (3, -4)
b. (6, -4)
c. (6, -8)
d. (-6, -4)
e. (-6, -8)

I already eliminated option c and e since based on the coefficient of y in the equation, the ordinate of the center must be -4. However, I don't know how to determine the abscissa since we need to determine the value of p first, in which we need to substitute the value of x and y while the only info I have is y = 0. How should I do this?
 

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circle touches the x-axis at one point $\implies x^2+px+9$ is a perfect square $\implies p$ is either +6 or -6 $\implies x$ is either -3 or +3,

choice (a) seems to be the only plausible fit ...

$x^2-6x+9 +y^2+8y+16=16$

$(x-3)^2 +(y+4)^2 =4^2$
 
Ah, thanks skeeter! That was faster than I thought...