Let ABCD be a square of side length $2$ units. $C_2$ is the circle through vertices $A,B,C,D$ and $C_1$ is the circle touching all the sides of the square ABCD. $L$ is a line through $A$
A circle touches the line $L$ and the circle $C_1$ externally such that both the circles are on the same side of the line, then the locus of centre of the circle is
My attempt is as follows:-
Let the center of circle which touches the line $L$ and circle $C_1$ externally be $C$
$$CC_1=r+r_1\tag{1}$$
Here $r$ is the radius of circle with center $C$ and and $r_1$ is the radius of fixed circle $C_1$ with radius $r_1$
As $L$ is tangent to the circle $CL=r\tag{2}$
Dividing $1$ and $2$
$$\dfrac{CC_1}{CL}=1+\dfrac{r_1}{r}$$
Actual answer is parabola but in that case $\dfrac{CC_1}{CL}=1$ as eccentricity of parabola is $1$, but here $1+\dfrac{r_1}{r}\ne1$
Is my assumption of considering $C_1$ and $L$ as fixed circle and fixed line wrong? Please help me in this.
HINT.
Consider a line $L'$ parallel to $L$ and at a distance $r_1$ from $L$, such that $CL'=r+r_1$.