polar tangents

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August 11th 2009, 12:42 AM
s_ingram

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polar tangents

Hi guys,

I am struggling with this polar equation problem:

The polar equation of a circle is $r = 4\cos\theta$ and two points $P(4,0)$ and $Q(2\sqrt{2}, \frac{\pi}{4})$ are found where the circle intersects the line $r = 2\sqrt{2}\sec(\frac{\pi}{4}-\theta)$ . Calculating P and Q was the first part of the problem. Now find the equations of the two half lines from the origin which are tangents to the circle which has PQ as diameter.

Maybe there is a way of doing this working directly with polar coordinates but I switch to cartesian and then convert back at the end. Perhaps this is not the best method - in which case I am open to suggestions.

In cartesian coordinates P is (4,0) and Q is (2,2) and the second circle with PQ as diameter can be seen to have radius $\sqrt 2$ and to pass through R(2,0) so I calculate it's equation to be $x^2 + y^2 -6x -2y + 8 = 0$ . See attachment. But I can't see a way to calculate the tangents from the origin! I don't know which points on the second circle will have tangents passing through the origin.

I tried implicit differentition to get a general gradient but I couldn't see how to use it. Can anyone help?
August 11th 2009, 01:17 AM
Grandad

Tangents to a circle in polar coordinates

Hello s_ingram

Quote:

Originally Posted by s_ingram

Hi guys,

I am struggling with this polar equation problem:

The polar equation of a circle is $r = 4\cos\theta$ and two points $P(4,0)$ and $Q(2\sqrt{2}, \frac{\pi}{4})$ are found where the circle intersects the line $r = 2\sqrt{2}\sec(\frac{\pi}{4}-\theta)$ . Calculating P and Q was the first part of the problem. Now find the equations of the two half lines from the origin which are tangents to the circle which has PQ as diameter.

Maybe there is a way of doing this working directly with polar coordinates but I switch to cartesian and then convert back at the end. Perhaps this is not the best method - in which case I am open to suggestions.

In cartesian coordinates P is (4,0) and Q is (2,2) and the second circle with PQ as diameter can be seen to have radius $\sqrt 2$ and to pass through R(2,0) so I calculate it's equation to be $x^2 + y^2 -6x -2y + 8 = 0$ . See attachment. But I can't see a way to calculate the tangents from the origin! I don't know which points on the second circle will have tangents passing through the origin.

I tried implicit differentition to get a general gradient but I couldn't see how to use it. Can anyone help?

I think you're overlooking some basic geometry here about tangents to a circle.

Clearly OQ is perpendicular to QP, and hence is the tangent at Q. Its polar equation is therefore $\theta = \frac{\pi}{4}$ .

And if the point of contact of the other tangent is T and the centre of the circle is C = (3, 1), then $\angle OTC = 90^o$ . So $\sin\angle COT = \frac{CT}{OC}= \frac{\sqrt2}{\sqrt{10}}$ . You can easily find the sine of the angle between OC and the x-axis. So you can work out the angle between OT and the x-axis, and hence its polar equation.

Grandad
August 11th 2009, 08:21 AM
s_ingram

You're right Grandad, I was overlooking the basic geometry because this is a polar equations question and I was expecting the answer to be linked to the polar transformations in some way. That's no excuse of course!

The other problem was that even though PQ is the diameter of a circle, I wasn't sure that a tangent at Q necessarily went through the origin. My diagram suggests it does of course, but how do you know that the tangent at Q passes though O?
August 11th 2009, 09:15 AM
Grandad

Hello s_ingram

Quote:

Originally Posted by s_ingram

... how do you know that the tangent at Q passes though O?

Because OQ has gradient 1 and QP has gradient -1. Therefore OQ and QP are perpendicular. But QP is a diameter, and therefore QO is a tangent, since a tangent is perpendicular to the radius through the point of contact.

Grandad
August 11th 2009, 10:37 AM
s_ingram

Ok Grandad, it's irrefutable!