Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre The line with equation $2x+y-5=0$ is a tangent to the circle with equation $(x-3)^2 + (y-p)^2=5$
Find the two possible values of $p$.

$$y=5-2x$$
Since this is a tangent, it touches the circle.
$$\begin{align}(x-3)^2+(y-p)^2&=5\\(x-3)^2+(5-2x-p)^2&=5\\x^2-6x+9+25+4x^2+p^2+2(-10x)+2(2px)+2(-5p)&=5\\5x^2+(4p-26)x+(p^2-10p+29)&=0\end{align}$$
We require that $$\boxed{b^2-4ac=0}$$
$$\begin{align}(4p-26)^2-20(p^2-10p+29)&=0\\4(2p-13)^2-20(p^2-10p+29)&=0\\4p^2-52p+169-5p^2+50p-145&=0\\p^2+2p-24&=0\\(p+6)(p-4)&=0\\p&=-6,4\end{align}$$
