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?
[Math] How to calculate radius when I know the tangent line length
circlesgeometrytriangles
Related Solutions
An identity that might prove useful in this problem is $$ \cot\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1-\cos(\theta)}=\frac{1+\cos(\theta)}{\sin(\theta)}\tag{1} $$ In $\mathbb{R}^3$, one usually uses the cross product to compute the sine of the angle between two vectors. However, one can use a two-dimensional analog of the cross product to do the same thing in $\mathbb{R}^2$.
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In the diagram above, $(x,y)\perp(y,-x)$ and so the $\color{#FF0000}{\text{red angle}}$ is complementary to the $\color{#00A000}{\text{green angle}}$. Thus, $$ \begin{align} \sin(\color{#FF0000}{\text{red angle}}) &=\cos(\color{#00A000}{\text{green angle}})\\[6pt] &=\frac{(u,v)\cdot(y,-x)}{|(u,v)||(y,-x)|}\\[6pt] &=\frac{uy-vx}{|(u,v)||(x,y)|}\tag{2} \end{align} $$ $uy-vx$ is the normal component of $(u,v,0)\times(x,y,0)$; thus, it is a two dimensional analog of the cross product, and we will denote it as $(u,v)\times(x,y)=uy-vx$.
Let $u_a=\frac{Q-P}{|Q-P|}$ and $u_b=\frac{R-P}{|R-P|}$, then since $$ |A-P|=|B-P|=r\cot\left(\frac{\theta}{2}\right) $$ we get $$ \begin{align} A &=P+ru_a\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_a\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ and $$ \begin{align} B &=P+ru_b\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_b\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ where we take the absolute value of $u_a\times u_b$ so that the circle is in the direction of $u_a$ and $u_b$.
You have equation of two tangents. STEP $1$:Draw perpendicular(normal) from centre to tangents it is equal to radius.since you have two tangents you will draw two perpendicular(normal).
STEP $2$ :so you will get two equation consists of coordinates of centre.
STEP $3$: Then solve those equation you will get coordinates of centre.If you dont know how to draw perpendicular let me know and give data I'll solve it.
Best Answer
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
$$a^2-2ax_0-r^2+16=0\Longrightarrow a=\frac{2x_0\pm\sqrt{4x_0^2+4r^2-64}}{2}=x_0\pm\sqrt{x_0^2+r^2-16}$$
Thus
$$OM=a=x_0+\sqrt{x_0+r^2-16}$$
and inputting in (**) above we get
$$r^2=2x_0^2+r^2-16+2x_0\sqrt{x_0^2+r^2-16}-16\Longrightarrow (32-2x_0^2)^2=4x_0^2(x_0^2+r^2-16)\Longrightarrow$$
$$1,024-128x_0^2+\rlap{\;/}4x_0^4=\rlap{\;/}4x_0^4+4r^2x_0^2-64x_0^2\Longrightarrow r^2=\frac{256-16x_0^2}{x_0^2}=16\frac{16-x_0^2}{x_0^2}=$$
$$=\left(\frac{4y_0}{x_0}\right)^2\Longrightarrow r=4\frac{y_0}{x_0}=4\tan\angle ROM$$