<s>[INST] 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 $p$.
 [/INST] The equation
$$(x-3)^2+(5-2x-p)^2=5$$
or
$$x^2+9-6x+4x^2+q^2-4qx=5$$
where $q=5-p $
must have only one solution.
$$5x^2-2x (2q+3)+4=0$$
$$\delta'=(2q+3)^2-20=0$$
thus
$$p=5-q=5-\frac {-3\pm 2\sqrt{5}}{2} $$
$$=\frac {13}{2}\pm \sqrt {5} $$
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