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?
[Math] Secant line and diameter of a circle
contest-mathgeometry
Related Solutions
I'll try to add a graph to help demonstrate that the maximal number of points of intersection between a circle and a rectangle such that the length of the rectangle is greater than the circle's diameter and its width is less than the diameter would be 6 such points of intersection: two points of intersection along each of the longest sides, and two points of intersection along one of the shorter sides. There is no way that there can be any points of intersection along the second of two shorter sides if the circle is intersecting the opposing side, since its diameter is less than the length of the rectangle.
Consider, for example, a circle of diameter 10 (radius 5) centered at the origin; hence its equation is $x^2 + y^2 = 25$. Consider a rectangle with vertices $(x_i, y_i)$ at $(-4, -8)$, $(4, -8)$, $(4, 4)$, $(-4, 4)$. Hence it's length (height) is $4 - (-8) = 12 > 10$, and its width is $4 - (-4) = 8 < 10$ (where 10 is the diameter of the circle). Then there are 2 points of intersection between the circle $x^2 + y^2 = 25$ and each of the line segments $y = 4$ ($-4 \leq x \leq 4$), $x = 4$ ($-8 \leq y \leq 4$), and $x = -4$ ($-8 \leq y \leq 4$), but no points of intersection between $x^2 + y^2 = 4$ and the rectangle's fourth side which lies on line $y = -8$ ($-4 \leq x \leq 4$). Solving for the points of intersection yields a total of 6 points of intersection of the circle and the rectangle: $(-4, -3), (-4, 3), (-3,4), (3, 4), (4, 3), (4, -3)$. ( If we move the circle vertically so it intersects the line $y = -8$, then it will no longer intersect the side along $y = 4$.
And there is no way a circle can intersect any given (straight) line in more than two points.
Set the center of the circle to be $\left(0,0\right)$. Then the set of points on the circle are those such that $x^2+y^2=r^2$ and the set of points on the line segment are solutions to $y=mx+b$ from some $m$ and $b$ (these are easily found given the two endpoints of the line segment). Substituting, we have \begin{equation} x^2+(mx+b)^2=r^2\Longleftrightarrow (m^2+1)x^2+2mbx+(b^2-r^2)=0 \end{equation} Then this is just a quadratic you can solve with the quadratic formula giving you $x$, and you can get $y$ by $y=mx+b$. However, this gives you two solutions; you must pick the right one (find which is between the two endpoint). Fortunately finding the angle after this is pretty easy: it's just simple trigonometry. If $(x,y)$ is the solution, $\tan^{-1}\left(\frac{y}{x}\right)$ gives you the angles measure counterclockwise from the $x$-axis. You asked for the angle from the $y$-axis, which is then $\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right).$
A worked example
I'll work out the example you have given. We have radius $r=\frac{3}{2}$, center $\left(3,4\right)$ and a line segment from $(3,3)$ to $(6,7)$. First we'll move the center to $(0,0)$. We do this by subtracting $(3,4)$ from each point. This gives us the center (as desired) of $(0,0)$, and a line segment from $(0,-1)$ to $(3,3)$. Then the line between the two points is $y=\frac{4}{3}x-1$. The equation for the circle is $x^2+y^2=\frac{9}{4}$. Substituting the equation for the line in, we have \begin{equation} x^2+\left(\frac{4}{3}x-1\right)^2=\frac{25}{9}x^2-\frac{8}{3}x+1=\frac{9}{4} \end{equation} which gives the quadratic \begin{equation} \frac{25}{9}x^2-\frac{8}{3}x-\frac{5}{4}=100x^2-96x-45=0 \end{equation} Then the quadratic equation gives $x=\frac{96\pm \sqrt{96^2+4\cdot 100\cdot 45}}{200}=-0.345,1.305$. We're going to want the solution with positive $x$, so $x=1.305$. Then $y=\frac{4}{3}(1.305)-1=0.740$. To find the desired angles, we merely take $90-\tan^{-1}\left(\frac{0.740}{1.305}\right)=90-29.56=60.44$. I'm not sure what the discrepancy between my answer and the one you provided is; it could be rounding errors, but it is more likely I just made a calculation error somewhere (please do point it out if you see it).
I know it seems like a lot, but the technique is relatively straight forward, the calculation is just some hard work. I hope this answer has shed some light on the process for you.
Best Answer
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.