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

#### Monoxdifly

##### Well-known member
The circle $$\displaystyle x^2+y^2+px+8y+9=0$$ 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?

#### skeeter

##### Well-known member
MHB Math Helper
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 ...

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

$(x-3)^2 +(y+4)^2 =4^2$

#### Monoxdifly

##### Well-known member
Ah, thanks skeeter! That was faster than I thought...