JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Circle

Let the circle with center on the x-axis be $x^2+y^2+\lambda x=0$, which has center $(-\lambda /2,0)$ and radius $|\lambda |/2$.

For the circle to be tangent to the parabola $y^2=4-x$ (i.e., $x=4-y^2$), substitute into the circle equation:
$x^2+(4-x)+\lambda x=0$
$x^2+x(\lambda -1)+4=0$

For tangency, discriminant = 0:
$(\lambda -1{)}^2-16=0$
$\lambda -1=\pm 4$
$\lambda =5$ or $\lambda =-3$

For the circle to be in the region (x ≥ 0), we need $\lambda =-3$.

The circle is $x^2+y^2-3x=0$ with radius $\frac{3}{2}=1.5$.