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The chord of contact of a circle of the tangents drawn from a point $P$ passes through $Q$. Length of the tangent from $P$ is $l$ and length of the tangent from $Q$ is $m$. Find $PQ$.

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closed as off-topic by zoli, Nick Peterson, John B, GNUSupporter 8964民主女神 地下教會, José Carlos Santos Jan 10 '18 at 23:09

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  • $\begingroup$ I request you to use Cartesian Co Ordinate Geometry $\endgroup$ – Arijit Sinha Jan 10 '18 at 16:02
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    $\begingroup$ What have you done so far? $\endgroup$ – Y. Forman Jan 10 '18 at 16:04
  • $\begingroup$ I got $\sqrt{l^2+m^2}$, but my proof is very ugly. $\endgroup$ – Michael Rozenberg Jan 10 '18 at 16:18
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Assuming that a solution exists, we can take for our circle $x^2+y^2=l^2$ and let $P=(l,l)$. The line on which $Q$ lies is given by $x+y=l$. Thus, the point $Q=(x,l-x)$ for some $x$ which is $<0$ or $>l$.

The length of a tangent line from $Q$ to our circle the second leg of a right triangle with one leg equal to $l$, and with hypotenuse $\sqrt{x^2+(l-x)^2}$. Thus we have:

$$m=\sqrt{x^2+(l-x)^2-l^2}=\sqrt{2x^2-2lx}.$$

Now, the distance $PQ$ is given by $$PQ=\sqrt{(x-l)^2+{((l-x)-l)^2}}=\sqrt{2x^2-2lx+l^2}.$$ Looking at both equations, we can put $PQ$ in terms of $m$:

$$PQ=\sqrt{m^2+l^2}$$

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  • $\begingroup$ This is the simplest method in my opinion. $\endgroup$ – Arijit Sinha Jan 10 '18 at 17:49
  • $\begingroup$ By choosing a particular radius and location for $P$, I have assumed that the solution does not depend on those, but only on the given information. You might need to provide an argument for why that's true. $\endgroup$ – G Tony Jacobs Jan 10 '18 at 17:59
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Let $AB$ be a given chord, $R$ be a radius of the circle and $O$ be a center of the circle.

Thus, $$AB=\frac{2Rl}{\sqrt{l^2+R^2}}$$ and $$QB\cdot QA=m^2$$ or $$\left(QA-\frac{2Rl}{\sqrt{l^2+R^2}}\right)QA=m^2,$$ which gives $$QA=\frac{lR}{\sqrt{l^2+R^2}}+\sqrt{\frac{l^2R^2}{l^2+R^2}+m^2}$$ and since $$\cos\measuredangle PAQ=\sin\measuredangle APO=\frac{R}{\sqrt{l^2+R^2}},$$ by law of cosines for $\Delta PAQ$ we obtain $$PQ=\sqrt{PA^2+QA^2-2PA\cdot PQ\cos\measuredangle PAQ}=$$ $$=\sqrt{l^2+\left(\tfrac{lR}{\sqrt{l^2+R^2}}+\sqrt{\tfrac{l^2R^2}{l^2+R^2}+m^2}\right)^2-2l\left(\tfrac{lR}{\sqrt{l^2+R^2}}+\sqrt{\tfrac{l^2R^2}{l^2+R^2}+m^2}\right)\tfrac{R}{\sqrt{l^2+R^2}}}=$$ $$=\sqrt{l^2+m^2}.$$

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By De La Hire's theorem, $Q\in\text{pol}(P)$ implies $P\in\text{pol}(Q)$.
By considering the powers of $P$ and $Q$ with respect to the given circle, we have that the circle centered at $P$ with radius $l$ and the circle centered at $Q$ with radius $m$ are orthogonal, hence $PQ^2 = l^2+m^2$ holds by the Pythagorean theorem.

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Related: How to construct three mutually orthogonal circles in stereographic projection?

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