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.
Best Answer
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.