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 $\displaystyle r = 4\cos\theta$ and two points $\displaystyle P(4,0)$ and $\displaystyle Q(2\sqrt{2}, \frac{\pi}{4})$ are found where the circle intersects the line $\displaystyle r = 2\sqrt{2}\sec(\frac{\pi}{4}-\theta)$. 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 $\displaystyle \sqrt 2$ and to pass through R(2,0) so I calculate it's equation to be $\displaystyle x^2 + y^2 -6x -2y + 8 = 0$. 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 $\displaystyle \theta = \frac{\pi}{4}$.

And if the point of contact of the other tangent is T and the centre of the circle is C = (3, 1), then $\displaystyle \angle OTC = 90^o$. So $\displaystyle \sin\angle COT = \frac{CT}{OC}= \frac{\sqrt2}{\sqrt{10}}$. 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