JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Complex Numbers

The first equation $|z-(4+3i)|=2$ represents a circle with center $(4,3)$ and radius $2$. The second equation $|z|+|z-4|=6$ represents an ellipse with foci at $(0,0)$ and $(4,0)$, where $2a=6$ so $a=3$, $c=2$, and $b^2=a^2-c^2=5$. The ellipse equation is $\frac{(x-2{)}^2}{9}+\frac{y^2}{5}=1$.

The lower extremity of the vertical diameter of the circle is at $(4,1)$. Substituting into the ellipse equation: $\frac{(4-2{)}^2}{9}+\frac{1}{5}=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}<1$. Since this point lies inside the ellipse, and the circle has radius 2, the circle intersects the ellipse at exactly 2 points.

Answer: (C)