The line y = mx + 1 cuts the parabola $y^2=4ax$ at two points P and Q. If equation  of circle with diameter as PQ is $x^2+y^2+18x+4y+5=0$ then a + m equal to 

$\begin{array}{c}by eleminating y from the givien equations {\left(mx+1\right)}^2=4ax\\ m^2{x}^2+2mx+1=4ax\\ m^2{x}^2+\left(2m-4a\right)x+1=0\\ \left(x-x_1\right)\left(x-x_2\right)=x^2+\frac{2x\left(m-2a\right)}{m^2}+\frac{1}{m^2}\mathrm{......}\left(1\right)\\ by eleminating x from the given equations y^2-4a\left(\frac{y-1}{m}\right)=0\\ \left(y-y_1\right)\left(y-y_2\right)=y^2-\frac{4ay}{m}+\frac{4a}{m}\mathrm{......}\left(2\right)\\ \text{Add the above two equations }\\ \left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=x^2+y^2+\frac{2x\left(m-2a\right)}{m^2}-\frac{4ay}{m}+\frac{4a}{m}+\frac{1}{m^2}\\ =x^2+y^2+18x+4y+5\\ m=\frac{1}{3},a=-\frac{1}{3}\end{array}$