Fixed and variable Circle question

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

• What are perpendicular normals to circle? – Maria Mazur Feb 17 at 10:53
• I mean line passing through centre and are mutually perpendicular – Samar Imam Zaidi Feb 17 at 10:55
• But then variable circle has fixed center – Maria Mazur Feb 17 at 10:56
• Normals are also tangent to $C_1$, and two perpendicular tangents form a square with respective radii. Hence... – 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 $$OX\cdot XY=(2\sqrt{2}-2)(2\sqrt{2}+2)=4$$