Two perpendicular normals to variable Circle are tangent to fixed circle $\ C_1$ with radius 2 and locus of centre of variable circle be the curve $\ C_2$, then find the product of maximum and minimum distance between the curves $\ C_1$& $\ C_2$.

My approach: Normal always pass through centre of circle . The normals are perpendicular to each other. B ut not able to proceed

  • $\begingroup$ What are perpendicular normals to circle? $\endgroup$ – Aqua Feb 17 at 10:53
  • $\begingroup$ I mean line passing through centre and are mutually perpendicular $\endgroup$ – Samar Imam Zaidi Feb 17 at 10:55
  • $\begingroup$ But then variable circle has fixed center $\endgroup$ – Aqua Feb 17 at 10:56
  • 1
    $\begingroup$ Normals are also tangent to $C_1$, and two perpendicular tangents form a square with respective radii. Hence... $\endgroup$ – Aretino Feb 17 at 11:24

We denote the center of variable circle by $O$, whose loci is the curve $C_2$. The center of the fixed circle $C_1$ is at $A$. Two perpendicular tangents to $C_1$ are passing through $O$ and touch the circle at $P$ and $Q$.


Since the circle's radius is perpendicular to the tangent line, $$AP\perp OP\quad\text{&}\quad AQ\perp OQ$$ and since $OP\perp OQ$ and $AP=AQ$ then $APOQ$ is a square whose side length is $2$.

This would imply that the distance between $A$ and $O$ is always a constant ($=2\sqrt{2}$), and since the point $A$ is fixed, then $O$ lies on a circle around $A$. In other words, $C_2$ is a circle with radius $2\sqrt{2}$.

To find the maximum and minimum distance between two circles, simply pass a line between their centers which in this case is any diameter of $C_2$. Evidently, the least distance between two points on $C_1$ and $C_2$ is equal to $OX$ and the farthest ones are $X$ and $Y$. Therefore, the product of maximum and minimum distances are $$OX\cdot XY=(2\sqrt{2}-2)(2\sqrt{2}+2)=4$$


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