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Two circles of radii in the ratio $1:2$ touch each other externally. The center of the small circle is '$c$' and that of the bigger circle is '$D$'.The point of contact is $A$. $\overline{PAQ}$ is a straight line where $P$ is on the smaller circle and $Q$ is on the bigger circle ($\overline{PAQ}$ does not pass through $c$). Angle $QAD = \alpha. $ Find the angle between the tangent at $Q$ to the bigger circle and the diameter of the smaller circle which passes through P (produced if necessary).

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    $\begingroup$ Hey @lokeshsangabattula, you asked liked 10 questions already. How about learning some $\LaTeX$ now? $\endgroup$ – Kaster Aug 22 '15 at 4:50
  • $\begingroup$ You have to give angle $QAD$ as $ \alpha $ or something. $\endgroup$ – Narasimham Aug 22 '15 at 4:53
  • $\begingroup$ @kaster sir i dont have latex on my pc $\endgroup$ – lokesh sangabattula Aug 22 '15 at 5:01
  • $\begingroup$ @narasimham okay then let QAD be alpha the how o find the required angle $\endgroup$ – lokesh sangabattula Aug 22 '15 at 5:04
  • $\begingroup$ Changed the title, hope ok $\endgroup$ – Narasimham Aug 22 '15 at 5:13
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When produced, let tangent at $Q$ and radius at $P$ cut at $Z$.

Angle at $ ZQA = \pi/2 - \alpha,ZPA = \alpha, $ because $ CP=CA$ isosceles triangle, and $ ZQA$ is the complement of $AQD.$

So angle at $Z$ is $$ \pi- (\pi/2 -\alpha + \alpha) = \pi/2. $$

So $ZP $ is perpendicular to $ZQ $ always independent of angle $\alpha$ value, or for that matter ratio of given circles as also coffeemath points out.

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  • $\begingroup$ Your answer came in while I was typing, and gives the same result as mine. (Perhaps I'll delete) Anyway +!. $\endgroup$ – coffeemath Aug 22 '15 at 5:59
  • $\begingroup$ You got it right, no need deleting. I dont know if multiple accepts are permissible. $\endgroup$ – Narasimham Aug 22 '15 at 6:13
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I may be wrong, but it seems the answer has nothing to do with the 2:1 ratio, and doesn't depend on $\alpha$. For reference imagine line $CAD$ is horizontal, and pick a point $E$ beyond $D$ and to its right, for angle reference.

Also imagine the point $P$ on the circle centered at $C$ to be somewhere below the line $CAD$, and on the diameter from $P$ to the circle centered at $C$ take another remote point $K$ somewhere above line $CAD$ again for angle reference.

Now if $\alpha$ is angle $QAD$ it is also angle $PAC$ since these angles are opposite each other in the intersecting lines $CAD$ and $PAC.$ In the circle centered at $C,$ the angle $PAC$ is also $\alpha$ since the equal radii make triangle $CPA$ isosceles. This means that angle $KCA$ is $2\alpha,$ since it is the exterior angle at vertex $C$ to triangle $CPA,$ so is the sum of the two remote interior angles, both $\alpha.$

In the circle centered at $D$ the same thing happens: Angle $DQA$ is also $\alpha$ (another isosceles triangle $DAQ$) and so the exterior angle $QDE$ is $2\alpha.$

But now we have angle $KCA$ equal to angle $QDE,$ both being $2\alpha.$ And this means lines $PCK$ and $QD$ are parallel. SO the tangent to the circle centered at $D$ formed at the point $Q$ is perpendicular to both line $DQ$ and the parallel line $PCK,$ and finally the intersection of these two lines will form an angle of 90 degrees.

This answer surprises me, and I'd appreciate it if anyone points out an error in the approach, etc, since the final result seems not to depend on either $\alpha$ or on the 2:1 radius ratio.

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    $\begingroup$ When two lines are parallel, a perpendicular to one of them makes it automatically perpendicular to the other.In this case it is a common perpendicular. $\endgroup$ – Narasimham Aug 22 '15 at 6:10

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