Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre

Question

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 $p$.

Answer 1

$$y=5-2x$$ 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 $$\boxed{b^2-4ac=0}$$ $$\begin{align}(4p-26)^2-20(p^2-10p+29)&=0\\4(2p-13)^2-20(p^2-10p+29)&=0\\4p^2-52p+169-5p^2+50p-145&=0\\p^2+2p-24&=0\\(p+6)(p-4)&=0\\p&=-6,4\end{align}$$

Answer 2

Use formula $$ d(C,l) = {|ax_0+by_0+c|\over \sqrt{a^2+b^2}}$$

where $C$ is a center of circle and $l$ given line.

In our case $d(C,l)=r =\sqrt{5}$ and $C= (3,p)$. So we have:

$$ \sqrt{5} = {|2\cdot 3+p-5|\over \sqrt{5}}$$

Finaly we have $|p+1|=5$ so $p=4$ or $p=-6$.

Answer 3

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 $$p=5-q=5-\frac {-3\pm 2\sqrt{5}}{2} $$ $$=\frac {13}{2}\pm \sqrt {5} $$