Locus of centre of circle

Without loss of generality, I choose the line to be $y=-1$ and the given circle to be centered on $(0;0)$ with radius $R$ . We're looking for circles with center $(x;y)$ and radius $r$ .

For both circles to intersect orthogonally, we must have $d^2=R^2+r^2$ . For the searched circle to touch the line, we must have $|y-(-1)|=r$ which is the same as $(y+1)^2=r^2$ .

As $d^2=x^2+y^2$ , we have then $x^2+y^2=R^2+(y+1)^2$ , that is $y=\frac{1}{2}(x^2-R^2-1)$ , which is a parabola .

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If we put $R=0$ in the formula, we have the known locus of points equidistant to a point $(0;0)$ and a line $y=-1$ , which is already well known to be a parabola. Moreover, we can see that the parabola for a given $R$ is just this same parabola with $R=0$ drawn a bit lower ( $-\frac{1}{2}R^2$ lower).

This is best solved through taking a special case, or else one unnecessarily has to apply the concept of determinant to find out it its parabola which is always a waste of time, i took the straight line as y=c and the circle randomnly, and saw that in the end the y^2 terms vanished but x^2 terms remained clearly a parabola