Eliminate $t$ from $h=\frac{3t^2-4t+1}{t^2+1}, k=\frac{4-2t}{t^2+1}$ 
Eliminate the parameter $t$ from $$h=\frac{3t^2-4t+1}{t^2+1}$$ $$k=\frac{4-2t}{t^2+1}$$

This is not the actual question. The question I encountered was:

Perpendicular is drawn from a fixed point $(3, 4)$ to a variable line which cuts positive x-axis at one unit distance from origin. Circle $S(x, y) = 0$ represents the locus of the foot of perpendicular drawn from point $(3, 4)$ to the variable line. If radius of $S(x, y) = 0$ is $\sqrt{λ_1}$ and length of tangent drawn to $S(x, y) = 0$ from origin is $\sqrt{λ_2}$ , then determine $\lambda_1$ and $\lambda_2$.

My attempt:
I attempted the original question by assuming the variable line as $x+ty-1=0$ and calculating the coordinates of foot of perpendicular $(h,k)$ on it from $(3,4)$. I got the coordinates $(h,k)$ as $$(h,k)\equiv \left(\frac{3t^2-4t+1}{t^2+1}, \frac{4-2t}{t^2+1}\right)$$
I am clueless after this point on how to eliminate $t$. Solving for $t$ from either of the equations and substituting it in the other is very cubersome and not a very efficient method because I'm preparing for a competitive exam. I verified using WolframAlpha that the locus is a circle but am unable to obtain it in the standard form.
I am looking for a relatively short method/trick for these type of locus. Thanks.
 A: The circle $S(x,y)=0$ passes through $A(3,0),B(1,4)$ and $C(3,4)$. (why?)
Since $\angle{ACB}=90^\circ$, the center $D$ of the circle is the midpoint of the line segment $AB$.
Let $E$ be the tangent point. Then, note that $\triangle{ODE}$ is a right triangle where $DE$ is the radius.
A: $$h-3=\frac{-4t-2}{t^2+1}\implies\frac{h-3}k=\frac{-4t-2}{4-2t}=\frac{2t+1}{t-2}=2+\frac5{t-2}$$
$$t=\frac5{(h-3)/k-2}+2=\frac{2h+k-6}{h-2k-3}$$
Then substitute this into the second equation to get an implicit form:
$$k=\frac{4-2(2h+k-6)/(h-2k-3)}{(2h+k-6)^2/(h-2k-3)^2+1}=\frac{-10k(h-2k-3)}{(2h+k-6)^2+(h-2k-3)^2}$$
$$(2h+k-6)^2+(h-2k-3)^2=-10(h-2k-3)$$
$$5h^2+5k^2-30h+45=-10h+20k+30$$
$$h^2+k^2-4h-4k+3=0$$
$$(h-2)^2+(k-2)^2=5$$
From this we get $\lambda_1=5$ and $\lambda_2=3$.
A: I hope you actually want to eliminate $t$ and form an equation of $h$ and $k$ if it is that then I hope it is a shortcut.
$$h+tk=1$$
$$\implies \frac{1-h}{k}=t$$
Now, $\require{cancel}$
\begin{align}&k=\frac{4-2\frac{1-h}{k}}{\frac{(1-h)^2}{k^2}+1}\
\implies& k=\frac{\frac{4k-2+2h}{k}}{\frac{h^2+1-2h+k^2}{k^2}}\
\implies &k=\frac{4k-2+2h}{\cancel{k}}\frac{k^{\cancel{2}}}{h^2+1-2h+k^2}\
\end{aliɡn}\
h²+k²-4h-4k+3=0
A: For the quick method:
By Thales’s theorem, we immediately have that the locus is the circle with $(3,4)$ and $(1,0)$ on a diameter. Thus, it has centre $(2,2)$ and radius $\sqrt 5$.
A: As an alternative to the other good answers, we can use the substitution $t=\tan\frac12\theta$. Then
$$h=3-\frac{4\tan\tfrac12\theta+2}{\sec^2\tfrac12\theta}\quad\text{and}\quad k=\frac{4-2\tan\tfrac12\theta}{\sec^2\tfrac12\theta}$$or
$\,h=2-\cos\theta-2\sin\theta\,$ and $\,k=2+2\cos\theta-\sin\theta.\,$ Hence
$$5\cos\theta=2k-h-2\quad\text{and}\quad5\sin\theta=6-2h-k,$$
giving $(2k-h-2)^2+(6-2h-k)^2=25$. This reduces to the circle
$$(h-2)^2+(k-2)^2=5,$$
which has centre $(2,2)$ and radius $\surd5$.
