Finding an equation of a circle with a given center and a tangent line

...call the radius squared as c for simplicity
Solve the equation simultaneously
Get a quadratic in x or y where c is contained in the coefficients
Look for repeated roots ...

...call the radius squared as c for simplicity
Solve the equation simultaneously
Get a quadratic in x or y where c is contained in the coefficients
Look for repeated roots ...

y=-2x+3

Sorry to keep bothering you but I did this:
(x-4)^2 + (-2x+3)^2=c
and all expanded and simplified gives me:
x^2-4x+5 which factorises to (x+1)(x-5)
I have two x values now but I'm not sure what to do next..

That's not quite right. You need to get the form $Ax^2+Bx+C=0$ then consider the discriminant and what the condition must be upon it for the line to be tangent (if the line is tangent, how many roots should this equation have? - ie points of intersection), where $C$ is in terms of $c$, and then use the condition on the discriminant to solve for $c$

In this case c represents r squared, I'm a bit confused.. I understand that if the line is a tangent there can only be one root, however I'm quite stuck

Alright, so you have $(x-4)^2+(-2x+3)^2=r^2$

Fully expand the LHS, collect all the common terms and get it in the form $Ax^2+Bx+C=0$ then apply the discriminant.

By fully expanding the LHS and collecting the like terms and simplifying it
I got x^2-4x+5= r^2

I searched to find out what a discriminant is and I still don't know, I'm really sorry

The centre of a circle is (4,0). The line 2x+y=3 is a tangent to the circle. Find the equation of the circle.

So I've racked my brains trying to handle this question, the first thing I did was
put the tangent into slope intercept form : y=-2x+3
and I know that the tangent is perpendicular to the radius so i figured out the equation of the radius where it meets the tangent which is: y=1/2 x-2.
Then, I tried to make the circle equation:
(x-4)^2+ y^2= r^2
however I don't know how to figure out what the radius is, help would be really appreciated
Here is the question along with the diagram

The included diagram makes me think you're supposed to take a more 'geometrical' approach, rather than substitutuing into the circle equation and looking for repeated roots.

The easiest way to solve this (assuming you know the formula for the distance between a point and a line) is to observe that the radius is simply the distance between the line y=-2x+3 and the point (4, 0).

Alternatively (and again, following what the diagram hints at): Find the equation of the other line in the diagram. That is, the line perpendicular to y=-2x+3 that passes through (4, 0). Find where this line intersects y=-2x+3. This gives you a point on the circumference and you can find the radius.

$(x^2-8x+16)+(4x^2-12x+9)=r^2 \Rightarrow 5x^2-20x+(25-r^2)=0$

The discriminant is $b^2-4ac$ for a quadratic in the form $ax^2+bx+c=0$