$26 \cdot 28$ The equations $x^{2}+y^{2}-16=0$ and $x^{2}-3 y+12=0$ have a common real solution, then the value of $y$ is
(A) only 4.
(B) $-7,4$.
(C) 0,4 .
(D) $y$ has no solution.
(E) all values of $y$.
(23rd American High School Mathematics Examination, 1972)
[Solution 1] Draw a circle with the origin as the center and 4 as the radius, and a parabola with a $y$-intercept of 4. From the graph, it is clear that they only intersect at the point $y=4$.
Therefore, the answer is $(A)$.
[Solution 2] Subtract the second equation from the first to get $y^{2}+3 y-28=0$, which gives $y=-7$ or 4.
When $y=4$, we find $x=0$, but for $y=-7$, there is no

real solution for $x$.
Therefore, the answer is $(A)$.