# Length of chord and sector of a circle

There is a line inside a circle that starts from a point on the diameter and reaches the circumference of the circle. It thus subtends an angle between itself and the diameter. It is as the following image :

With the radius, r, the offset from center, a, and the angle subtended by the line with the diameter, b, known, how can we find :

1. The length of the line, x
2. The length of the sector, y
• what's been tried ? also I think stackechange has it's own image site. – user451844 Jul 14 '17 at 19:05
• $r^2=a^2+x^2-2ax \cos b$ – N74 Jul 14 '17 at 19:12

For $x$, join the center of the circle to the intersection of the "inside" line (marked with length $x$) and the circle. A triangle thus formed has side lengths $a,r$ and $x$ and an internal angle $b$. Now apply cosine rule to solve for $x$. $$\cos b = \frac{a^2+x^2-r^2}{2ax}$$
After this construction and finding $x$ it will be easy to get $y$. Can you take it from here?
• So you are solving a quadratic equation of the form $ax^2+bx+c=0$. The solutions to that are given by $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. In your case $a=1,b=-6$ and $c=-21.9$. – Anurag A Jul 14 '17 at 19:22
• @Ashley When angle $b=90^{\circ}$, then $\cos b=0$. So the equation reduces to $3^2+x^2-5^2=0$, which gives $x=4$. I think you might have plugged in the incorrect value somewhere. – Anurag A Jul 15 '17 at 13:04