Tangents to Circle

Question

Find the equation of the tangent at the point (acosØ,asinØ) on the circle x^2+y^2=a^2. Tangents are drawn from the point (2a,2a) to the above circle. If the coordinates of the point of contact be denoted by (acosØ,asinØ) prove that Tan(Ø/2)= 1/3[2±√7]...(1).

The answer to the first part is xcosØ+ysinØ=a. I tried to solve the second part by using the formula for two tangents to a circle from a point outside it-m^2(x1^2-a^2)-2mx1y1+(y1^2-a^2). Substituting (2a,2a) for (x1,y1) results in m=1/3[4±√7]. I now have the slopes of the two tangents and relating this back to (1), indicates that these slopes are correct.

Now the problem! I have been unable to reproduce the formula (1) above, nor have I found any reason for using Tan(Ø/2) in the calculations. I have tried using the polar coordinates to calculate the tangent\normal slopes with (2a,2a)"m1" and (0,0)"m2" and use the formula m1xm2=-1 to try to get to (1) above, without success. I also tried using the answer from the first part to solve the second part by comparing coefficients, again without success. Any help would be greatly appreciated.

score 1 · Accepted Answer · answered Mar 20 '13 at 14:49

1

You are almost done!

satisfy(2a,2a) in general tangent equation and you get ==> $$ sin(\theta) +cos(\theta) = 1/2 $$ put $$ sin(\theta)=2tan(\theta/2)/[1+tan^2(\theta/2)]$$ and $$ cos(\theta)=[1-tan^2(\theta/2)]/[1+tan^2(\theta/2)] $$

you get a quadratic equation in $$tan(\theta/2)$$ which can be solved to get the desired result.

answered Mar 20 '13 at 14:49

ABC

6,034

Thanks, the trig. identities were unknown to me, I have added them to my list! – twa14 Mar 20 '13 at 15:59
these identities are very helpful in integration too. – ABC Mar 20 '13 at 16:10
How do I reward you? – twa14 Mar 20 '13 at 16:27
@twa14 Just click on the small tick symbol under the vote count of my answer. even u will be awarded for accepting! – ABC Mar 20 '13 at 16:30

Tangents to Circle

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