Consider a family of lines (4a + 3)x -(a+1)y -(2a+1) = 0 
Consider a family of lines $(4a + 3)x -(a+1)y -(2a+1) = 0$ where $a \in R$. The locus of the foot of the perpendicular from the origin of each member of this family is?

My Attempt:
Let the foot of the perpendicular from the origin on these lines be $(\alpha,\beta)$
Applying the formula of the foot of the perpendicular, I get
$$\alpha = \frac{8a^2 + 10a +3}{17a^2+26a+10}$$
$$\beta = -\biggl(\frac{2a^2+3a+1}{17a^2+26a+10}\biggr)$$
Now, I know that I need to eliminate '$a$' in order to get the locus. But, how am I supposed to do that? It's like quadratic on the top and bottom. Using the quadratic formula would be messy. 
Any help would be appreciated.

One approach: write the two equations as polynomials in $a$ and compute their resultant. To wit, rewrite the equations $$x = {8a^2+10a+3 \over 17a^2+26a+10} \\ y = -{2a^2+3a+1 \over 17a^2+26a+10}$$ as $$(17a^2+26a+10)x-(8a^2+10a+3) = (17x-8)a^2+(26x-10)a+(10x-3) = 0 \\ 
(17a^2+26a+10)y +(2a^2+3a+1) = (17y+2)a^2+(26y+3)a+(10y+1) = 0.$$ Now form the Sylvester matrix of these two polynomials and compute its determinant (a symbolic algebra program is helpful here): $$\det\begin{bmatrix}17x-8&26x-10&10x-3&0 \\ 0&17x-8&26x-10&10x-3 \\ 17y+2&26y+3&10y+1&0 \\ 0&17y+2&26y+3&10y+1 \end{bmatrix} = 5(x^2-x+y^2-2y).$$ The resulting implicit Cartesian equation is therefore $$x^2+y^2-x-2y=0,$$ which is the equation of a circle of radius $\frac{\sqrt5}2$ centered at $\left(\frac12,1\right)$.
