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 $C$


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 $2$


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.

  • $\begingroup$ yeah we can leave that, it has no importance. $\endgroup$ Feb 8 '20 at 9:48
  • $\begingroup$ FYI - $C_1$ is the "inscribed" circle, and $C_2$ is the "circumscribed" circle. $\endgroup$ Feb 8 '20 at 15:24
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    $\begingroup$ Your mistake is in assuming that the center of $C_1$ is the focus and $L$ is the directrix. Since the ratio of the distances is not constant, but varies with $r$, they cannot be the focus and directrix of a conic section. (That doesn't say the curve isn't a conic section - just that these two do not fit the roles of focus and directrix.) $\endgroup$ Feb 8 '20 at 15:39


Consider a line $L'$ parallel to $L$ and at a distance $r_1$ from $L$, such that $CL'=r+r_1$.


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