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For my math homework, I was asked this question: The tangent lines from O hit a circle with center M and radius r in R and S. Calculate r.

-The length of OR and OS is 4

How do I calculate the radius r of circle M from this?

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OR is perpendicular to RM and so is OS to MS. Also notice that ON = NM and BM = r. What can you say about angle ROS? – Dilawar Nov 17 '12 at 13:36
The only thing you know is $OR=OS=4$? That is definitely not enough to determine $r$. Are you sure you're not omitting something from the homework? – Rahul Nov 17 '12 at 14:09
Hmm... I see... The math book wasn't very clear on this matter. I guess that means I have to re-use the M from the previous question (OM = 8). This makes DonAntonio's answer valid. Thanks for your replies. – Qqwy Nov 17 '12 at 14:21
Ah, too bad I didn't read the above before I added that stuff to my answer. Anyway, knowing some angle there in the origin could help, too. – DonAntonio Nov 17 '12 at 15:01
Heh, Thanks a lot for your second answer. It's nice to know how to calculate it when you know the angle in the origin. – Qqwy Nov 17 '12 at 15:14
up vote 1 down vote accepted

I'm assuming you know the center $\,M\,$ , so we get that:

$$\angle ORM=90^\circ\Longrightarrow r^2=MR^2=OM^2-OR^2$$

applying Pythagoras Theorem.

Added: Let $\,R=(x_0,y_0)\,\,,\,\,S=(x_0,-y_0)\,$ , so that if $\,M=(a,0)\,$ we then have:

$$\begin{align*} x_0^2+y_0^2=&16\\(x_0-a)^2+y_0^2=&r^2\end{align*}$$

From the second equation we get

$$a^2-2ax_0-r^2+16=0\Longrightarrow a=\frac{2x_0\pm\sqrt{4x_0^2+4r^2-64}}{2}=x_0\pm\sqrt{x_0^2+r^2-16}$$



and inputting in (**) above we get

$$r^2=2x_0^2+r^2-16+2x_0\sqrt{x_0^2+r^2-16}-16\Longrightarrow (32-2x_0^2)^2=4x_0^2(x_0^2+r^2-16)\Longrightarrow$$

$$1,024-128x_0^2+\rlap{\;/}4x_0^4=\rlap{\;/}4x_0^4+4r^2x_0^2-64x_0^2\Longrightarrow r^2=\frac{256-16x_0^2}{x_0^2}=16\frac{16-x_0^2}{x_0^2}=$$

$$=\left(\frac{4y_0}{x_0}\right)^2\Longrightarrow r=4\frac{y_0}{x_0}=4\tan\angle ROM$$

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I added some stuff yo my answer. Hopefully that will help you out. – DonAntonio Nov 17 '12 at 14:57

protected by Bookend Apr 13 at 21:03

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