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Step-by-Step Solution
1. Assume the center of the circle
Let the center of the required circle be at the point
$ (h, k) $ in the $xy$-plane.
2. Use the chord condition on the x-axis
Since the circle cuts a chord of length $4a$ on the x-axis,
we can take the endpoints of the chord to be $(-2a,\,0)$ and $(2a,\,0)$
without loss of generality.
The perpendicular distance from the center $(h, k)$
to the chord (the x-axis) is $|k|$.
For a chord of length $4a$, the relationship with the circle’s radius $r$ is:
$4a \;=\; 2\sqrt{\,r^2 \;-\; k^2}\;\;\Longrightarrow\;\;r^2 \;=\; k^2 + 4a^2.$
3. Impose the condition that the circle passes through (0, 2b)
The circle passes through the point $(0, 2b)$. Hence, the distance
from $(h, k)$ to $(0, 2b)$ is also the radius $r$:
$ \sqrt{\,(0 - h)^2 + (2b - k)^2}\;=\;\sqrt{\,r^2}\;=\;\sqrt{\,k^2 + 4a^2}. $
Squaring both sides gives:
$h^2 \;+\;(2b - k)^2 \;=\; k^2 + 4a^2.$
Expand $(2b - k)^2$ to get:
$h^2 \;+\; (4b^2 \;-\; 4bk + k^2) \;=\; k^2 + 4a^2$
$h^2 \;+\; 4b^2 \;-\; 4bk \;+\; k^2 \;=\; k^2 + 4a^2$
Simplify:
$h^2 \;+\; 4b^2 \;-\; 4bk \;=\; 4a^2$
$h^2 \;=\; 4a^2 \;+\; 4bk \;-\; 4b^2.$
4. Express the locus in a conic section form
Rearrange the above equation:
$h^2 \;=\; 4\bigl[bk \;-\; b^2 \;+\; a^2\bigr].$
If we identify $h$ as the “$x$” variable and $k$ as the “$y$” variable
of the locus, we can write:
$x^2 \;=\; 4\bigl[b\,y \;-\; b^2 \;+\; a^2\bigr].$
This represents a parabola in the $xy$-plane (with $x$ playing the role
of $h$ and $y$ playing the role of $k$).
5. Conclusion
Therefore, the locus of the center of the circle is
a parabola.
