# Finding a distance between a point in a circle from the center.

Given a diagram like this,

Where $$O$$ is the center and $$OA = \sqrt{50}$$, $$AB = 6$$, and $$BC = 2$$. The question was to find the length of $$OB$$. $$\angle ABC = 90^o$$

What I've done is so far:

I made the triangle $$ABC$$ and named $$\angle BAC = \alpha$$ . By trigonometry, I have the values for $$\sin{\alpha}$$ and $$\cos{\alpha}$$. I get $$\cos{\alpha}=\frac{6}{\sqrt{40}}$$.

Then I made the triangle $$OCA$$ and named $$\angle OAB = \beta$$ so $$\angle OAC = \alpha + \beta$$. By using the cosinus rule, I have $$\cos(\alpha + \beta) = \frac{1}{\sqrt{5}}$$.

Using the formula, $$\cos(\alpha + \beta) = \cos{\alpha}.\cos{\beta} - \sin{\alpha}.\sin{\beta}$$ and making $$\sin{\beta} = \sqrt{1 -\cos^2{\beta}}$$ I finally get that $$\cos{\angle OAB} = \frac{1}{\sqrt{2}}$$.

Finally, by using the cosinus rule on the triangle $$AOB$$ I get $$OB = \sqrt{26}$$.

My only problem is this takes me way too long! I am interested in a quicker way to do this (i.e. I now know that $$\angle OAB = 45^o$$ from trigonometry, but is there a quicker way to recognize it?)

• The angle OAB cannot be $45$ degrees. – Mark Sapir Jul 15 '20 at 15:01
• Are we also given that $\angle ABC$ is right angle? – Jaap Scherphuis Jul 15 '20 at 15:09
• @JaapScherphuis yeah! I forgot, sorry. – aco Jul 15 '20 at 15:41

Assuming $$\angle ABC=90^o$$ is given.

You can get there slightly quicker:
By Pythagoras, $$|AC|=\sqrt{40}$$.
$$OAC$$ is isosceles, with $$|OA|=|OC|=\sqrt{50}$$.
You can then immediately get $$\cos(\angle OAC)=\frac{|AC|/2}{|OA|} = \frac{\sqrt{40}/2}{\sqrt{50}}= \frac{1}{\sqrt{5}}$$.
I don't yet see a way to shortcut the rest.

You could do it completely differently, by algebra. Use a coordinate system, centred on $$B$$, and let $$O$$ be the point $$(x,y)$$. Then we get two equations from the fact that $$|OA|=|OC|=\sqrt{50}$$.

$$x^2+(6-y)^2=50\\ (2-x)^2+y^2=50$$

There are fairly easily solved to give $$y=1$$, $$x=-5$$, from which you get $$|OB|=\sqrt{26}$$.

Refer to the figure:

$$\hspace{4cm}$$

From the right triangle $$ACD$$: $$CD=\sqrt{AD^2-AC^2}=4\sqrt{10}$$.

From similarity of right triangles $$ABC$$ and $$CDE$$: $$\frac{CE}{AB}=\frac{CD}{AC}\Rightarrow CE=12\\ DE=\sqrt{CD^2-CE^2}=4=BF\\ BE=CE-BC=12-2=10=DF=AF$$ Hence, $$\angle DAF=45^\circ=\angle OAB$$, indeed.

Finally, from the cosine theorem for $$\triangle AOB$$: \begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\ &=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\ &=\sqrt{26}.\end{align}

A slight variation of the solution

Note that $$R$$ is circumradius of $$\triangle ADC$$,

\begin{align} |CD|&=2R\sin\alpha=2\sqrt5 ,\\ |BD|&=\sqrt{|CD|^2-a^2}=4 ,\\ |AD|&=c+BD=10 . \end{align}

By the Stewart’s Theorem for $$\triangle AOD$$,

\begin{align} |OD|^2\cdot c+|OA|^2\cdot |BD| -|AD|\cdot(|OB|^2+c\cdot |BD|) &=0 ,\\ |AD|\cdot ( R^2 -|OB|^2-c\cdot |BD|) &=0 , \end{align}

\begin{align} |OB|^2&= R^2-c\cdot |BD| \\ &=50-6\cdot4 =26 . \end{align}