# Locus of a point $P$ such that the tangents to 2 concentric circles issued from $P$ are orthogonal

If the tangent from a point P to the circle $$x^2+y^2=1$$ is perpendicular to the tangent from P to the circle $$x^2+y^2=3$$, then the locus of P is,

So this is what the question means diagrammatically.

Since the question said PA and PB are perpendicular, let PA=$$m_1$$ & PB=$$m_2$$, then $$m_1 m_2=-1$$.

Preferably I would want a more geometrical approach

• The diagram does not appear to relate to the actual positions of the circles, since one should be inside the other... Commented Sep 24, 2020 at 0:48
• @abiessu edited!
– rash
Commented Sep 24, 2020 at 0:58
• from a point $P$ to any circle $\mathcal{C}$, you can draw 2 tangents
– L F
Commented Sep 24, 2020 at 1:00
• i.sstatic.net/J6R7w.png
– L F
Commented Sep 24, 2020 at 1:02

Since these circles are concentric, we only need to consider the tangents $$x=-1$$ and $$y=\sqrt{3}$$. By applying the Pythagorean Theorem, we find this $$P$$ is two units away from the origin. Because the setup has rotational symmetry, the locus of $$P$$ is the circle centered at the origin that passes through $$(-1,\sqrt{3})$$, i.e. $$x^2+y^2=4.$$

A characteristic property of a tangent to a circle with center $$O$$ with point of tangency $$T$$ is orthogonal to $$OT$$.

Therefore, a point $$P$$ is a solution, if and only there exist (tangency) points $$A$$, resp. $$B$$ on the smallest, resp. the largest circle such that quadrilateral $$OBPA$$ is a rectangle. As the sides $$OA$$ and $$OB$$ have fixed length $$1$$ and $$\sqrt{3}$$, the diagonal of the rectangle $$OP$$ has a fixed length $$\sqrt{1^2+\sqrt{3}^2}=2$$.

Conclusion: The locus of point $$P$$ is the circle with center $$O$$ and radius $$2$$.

Remark: A very similar question can be found here.