Mathematics - Circle Question with Solution | TestHub

$(x-p)(x-\alpha )+(y-g)(y-\beta )=0$ (or) $x^2+y^2-(p+\alpha )x-(g+\beta )y+p\alpha +g\beta =0\to (1)$ Put $y=0$, we get $x^2-(p+\alpha )x+p\alpha +g\beta =0\to (2)$ $\therefore \Rightarrow$ Locus of $B(\alpha ,\beta )$ is $(p-x{)}^2=4gy$

$$(x-p{)}^2=4gy$$