calculuscirclesgeometrytangent line
given circle:
$$x^2 + y^2 – 2x – 4y – 4 = 0,$$
given line which is parallel to the tangent:
$$3x – 4y – 1 = 0,$$
I found out the center and radius of the given circle which came out to be $(1,2)$ and $3$ respectively.
but how do i find out the equation of the tangent if i don't know the point where it touches the circle? i know two parallel lines have the same slope. so the tangent and that line should have slope of $3/4$. but that still doesn't give the points to form the equation
A useful technique for this situation which avoids using calculus goes like this:
Find the equation of a line through the point $(4,-3)$ with gradient $k$, and then use this (linear) equation to substitute into the equation of the circle to get a quadratic. The two roots of this quadratic would give the 2 points of intersection of the line with the circle, but in our case, we want to know the values of $k$ for which the quadratic has a double root (i.e. the 2 points of intersection are actually coincident). Use the standard condition ($b^2=4ac$) for a double root of a quadratic.
In your case, the equation of the line through $(4,-3)$ with gradient $k$ is $y+3 = k(x-4)$, or $y+7 = k(x-4)+4$. Substituting this into the equation of the circle gives:
$$(x+2)^2 + (k(x-4)+4))^2 = 25$$
So you need to rearrange this quadratic in $x$ and find out which vales of $k$ give double roots.
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
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: