A secant line incident to a circle at points $A$ and $C$ intersects the circle's diameter at point $B$ with a $45^\circ$ angle. If the length of $AB$ is $1$ and the length of $BC$ is $7$, then what is the circle's radius?
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$\begingroup$The above diagram is almost self explanatory. The perpendicular bisector of chord AC passes through the center O of the circle. Since the diameter line makes an angle of 45 degrees with AC, angle MBO is 45 degrees and so triangle MBO is isosceles. Hence |MO|=3, |AM|=4 and by Pythagoras, r = |OA| =5.
$\endgroup$ $\begingroup$We can set up a system of equations for $BE$ and $BD$, then that will give us the radius, which is $$r=\frac{BE+BD}{2}$$ First, by the chord-chord theorem (AKA the power chord theorem), we can write $$AB\times BC=BE\times BD$$ $$\Rightarrow BE\times BD=7\quad\quad\quad(1)$$ Next, we can use the cosine law in $\triangle BCO$ (note that $CO$ is also a radius): $$CO^2=BC^2+BO^2-2\times BC\times BO\times \cos \angle CBO$$ $$\Rightarrow r^2=49+(BE-r)^2-14(BE-r)\cos(45)$$ $$\Rightarrow r^2=49+(BE-r)^2-7\sqrt{2}(BE-r)$$ $$\Rightarrow \left(\frac{BE+BD}{2}\right)^2=49+\left(\frac{BE-BD}{2}\right)^2-7\sqrt2\left(\frac{BE-BD}{2}\right)$$ $$\dots$$ $$\Longrightarrow BE\times BD+\frac{7\sqrt2 BE}{2}-\frac{7\sqrt2 BD}{2}=49$$ $$\Rightarrow \left(BE-\frac{7\sqrt2}{2}\right)\left(BD+\frac{7\sqrt2}{2}\right)=\frac{49}{2}\quad\quad\quad(2)$$ Can you continue the algebra?
$\endgroup$ $\begingroup$First step: Let $O$ be the centre of the circle. Apply the law of sines to $\triangle OBA$ and $\triangle OBC$.
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