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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.

enter image description here

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

I am stuck here and I do not know how to proceed. Please help.

Preferably I would want a more geometrical approach

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

2 Answers 2

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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. $x^2+y^2=4.$

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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.

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