Calculate the Length of a Circle's Radius

[neilparker62]

Ok.

Using the figure below I have using the Law of Cosine:
$$
\begin{aligned}
a^2&=b^2+c^2-2bc\cos(\theta)\\
a^2&=2r^2-2r^2\cos{\theta} \\
a^2&=2r^2(1-\cos{\theta}) \\
\end{aligned}
$$

cos(θ) = cos(90 + δ) = -sin(δ) where sin(δ) = (10-r) / r.

or produce the radius forming a diameter 2r. Then:

a/2r = cos(β)=sin(α)=10/a

 

[kuruman]

I took upon myself to summarize a solution for the benefit of anyone you might look at this thread and try to sift through the various approaches for getting to the answer. I think the most straightforward strategy has been outlined in post #14.
1. Find the equation of the line connecting points A {0, 10} and B {3,0} that lie on the circle. Can be done by inspection.
Answer: $y=-\frac{10}{3}x+10.$
2. Find the equation of the perpendicular bisector of segment AB. The slope is the negative inverse of the slope of AB, $m=\frac{3}{10}$. It must pass through the midpoint of AB, point C {$\dfrac{3}{2},5$}. To find the intercept $b$, solve $$5=\frac{3}{10}\times \frac{3}{2}+b ~\rightarrow~b=\frac{91}{20}.$$Answer: $y=\dfrac{3}{10}x+\dfrac{91}{20}.$
3. Find the intersection of the perpendicular bisector and the line $x=3$. Just plug in $x=3$ in the equation found in step 2.
$$y=\frac{3}{10}\times 3+\frac{91}{20}=\frac{109}{20}=5.45.$$Answer: The center of the circle is at point O {3, 5.45} and the radius of the circle is $R=5.45.$

 

[chwala]

Thanks Kuruman, I will look at it in depth later...