circlestangent line
So I have a circle: $(x-2)^2 + (y-2)^2 = 25$ and I have a tangent line to this circle, with a slope of $m= -3/4$.
I have to find the equation of the tangent line, so I know the radius of the circle is $r = 5$ and I wrote the equation of the tangent line as:
$$y = -3/4x + h$$
So now I have to find $y, x$ and $h$, but I don't know if I can just replace $x$ and $y$ with the center points? Or do I have to find the point-line distance (and why?)
You've found the radius and centre, then let the tangent line be $3x-4y+c=0$.
The distance of the tangent from the centre is
\begin{align*} \left| \frac{3(1)-4(2)+c}{\sqrt{3^2+4^2}} \right| &= 3 \\ \left( \frac{c-5}{5} \right)^2 &= 3^2 \\ (c-5)^2-15^2 &= 0 \\ (c-20)(c+10) &= 0 \\ c &= 20 \quad \text{or} \quad -10 \end{align*}
The required tangents are:
$$3x-4y+20=0 \quad \text{or} \quad 3x-4y-10=0$$
We have $(R-r)^2 = (a-r)^2 +b^2$ and hence $$r = \frac{R^2-a^2-b^2}{2(R-a)}$$
Best Answer
The center of the circle can be figured out by the given equation of the circle and is the point : $$C(2,2)$$
But then, if the given line is tangent to your circle, it means that the distance from the center of the circle should be exactly $5$. Manipulating the line equation you have derived, we can yield : $4y + 3x + k = 0$ and then by solving the distance formula, you can yield the exact equation (there will be 2 parallel and diametrically opposite equations thus two tangent lines) :
$$ \left|\frac{Ax_0 + By_0 + Γ}{\sqrt{A^2 +B^2}} \right| = d(P,ε) \Rightarrow \left|\frac{4\cdot 2 + 3 \cdot 2 + k}{\sqrt{4^2+3^2}}\right| = 5$$
$$\Leftrightarrow$$
$$|8 + 6 + k| = 25 \Leftrightarrow \dots$$
Another approach would be substituting the line equation for $x$ and $y$ into your circle's equation and then demanding the equation to have a unique solution, since a tangent line will only have one common point with a circle.
Note : This only works for the case of the circle, when a tangent line can never have $2$ common points. This is not the case for other curves though.