Problem :

Let $P$ be a moving point such that if $PA$ and $PB$ are two tangents drawn from $P$ to the circle $x^2+y^2=1$ ( $A$, $B$ being the points of contact) , then $\angle AOB = 60^{\circ}$, where $O$ is origin.Then find the locus of $P$.

My approach :

We know that tangents drawn from same point to the same circle are congruent. Also, line joining the centre of this circle to the point of tangency is perpendicular i.e. angle formed by tangent and radius in $90^{\circ}$

OA and OB = 1 unit.

But I am unable to proceed further , I know that locus is a circle but unable to find the radius of this circle which is OP. Please guide how to find the radius of this circle. Thanks...

  • 1
    $\begingroup$ Have you tried investigating the triangle $OPA$, for which the observations you have made will tell you all the angles, and you have the length of the side $OA$? $\endgroup$ – Mark Bennet Sep 30 '13 at 17:00
  • $\begingroup$ OA and OB is equal to 1. I am talking about the radius OP.. ? $\endgroup$ – sultan Sep 30 '13 at 17:16
  • $\begingroup$ If you know the angles of a (right-angled) triangle and one side, you can find the other sides. From $AO$ and the angles you can find $OP$. $\endgroup$ – Mark Bennet Sep 30 '13 at 17:21

First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.

From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.

Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:

$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$


We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:

$$\angle BAP = 30^{\circ}$$

Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:

$$\tan \angle BAP = \frac{PD}{AD}$$ $$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$ $$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$

$$ PD = \frac{1}{2\sqrt{3}}$$

Now we have: $PO = OD + PD$, i.e.:

$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{4}{2\sqrt{3}}$$ $$PO = \frac{2}{\sqrt{3}} \approx 1.154$$

So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:

$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$ $$x^2 + y^2 = \frac 43$$

So the locus of the point $P$ is:

$$x^2 + y^2 = \frac 43$$

  • $\begingroup$ that was genuinely a perfect explanation of this problem... I am very greatful to you for this... Thanks a ton. $\endgroup$ – sultan Oct 1 '13 at 3:02
  • $\begingroup$ Actually this is just over complicated solution, because we can consider the triangle $AOP$, which is right angle, with angles $30^{\circ}$ and $60^{\circ}$, and then just apply trigonometry. That would have been a simplier answer, but ultimately we would have reached the same solution. $\endgroup$ – Stefan4024 Oct 1 '13 at 11:41

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