Algebra Challenge – 2 ⭐⭐

Solution: The orthocenter $P$ is the point where line $AP$ and $CP$ intersects with either other. Assume $C=(X_c, Y_c)$, we have $X_c^2+Y_c^2=4$. Therefore $$Y_c=\pm\sqrt{4-X_c^2}\tag{1}$$ And the equation of line $CP$ is $$x=X_c\tag{2}$$ Because the slop of $BC$ is $\dfrac{Y_c}{X_c-1}$, the equation of line $AP$ is $$y=-\dfrac{X_c-1}{Y_c}(x+1)\tag{3}$$ By substituting $X_c$ and $Y_c$ in equation (3) with $(1)$ and $(2)$, we have $$y=\dfrac{1-x^2}{\pm\sqrt{4-x^2}}$$ Squaring both side of the above equation, the locus of point $P$ is the following function: $$\boxed{y^2=\dfrac{(1-x^2)^2}{4-x^2}}$$