# Drawing vertical line through a circle to find for what x does that vertical line div

• May 19th 2013, 08:06 PM
pandakrap
Drawing vertical line through a circle to find for what x does that vertical line div
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Hi all, I would appreciate some help with the problem:

The problem state: "A building design consultant is going to make a 14 foot diameter windowwith a vertical cut through the glass so that 2/3 of the glass can be tinted. Wheres should this cut be made?"

As you can see in the above attachment, I was able to conclude into some sort of equation, but when I solve that equation to 49*pi/3 via wolfmath, I got about 6.83 = x, which does not make sense because the radius is 7 and that wouldn't be 1/3 if you draw the vertical line through it to divide the circle.

If you could help me out, I would really appreciate it.

Thank you,
• May 19th 2013, 09:09 PM
Prove It
Re: Drawing vertical line through a circle to find for what x does that vertical line
First note that the area of this circle would be \displaystyle \displaystyle \begin{align*} \pi \cdot 7^2 = 49\,\pi \textrm{ units}\, ^2 \end{align*}

If we place the circle on a set of axes centred at the origin, it has equation \displaystyle \displaystyle \begin{align*} x^2 + y^2 = 7^2 \end{align*} then draw in the vertical line \displaystyle \displaystyle \begin{align*} x = a \end{align*} somewhere to the right of the y-axis, we wish to know what value of x this is so that it splits the circle into areas in the ratio 2:1. So that means the area bounded between this line and the circle on the right will be \displaystyle \displaystyle \begin{align*} \frac{49 \, \pi}{3} \,\textrm{units}\,^2 \end{align*}, and since this is evenly distributed above and below the x-axis, the area above the x-axis is \displaystyle \displaystyle \begin{align*} \frac{49 \, \pi }{6} \, \textrm{units} \, ^2 \end{align*}.

So we can set up an integral for the area bounded by the circle, the x-axis and the vertical line \displaystyle \displaystyle \begin{align*} x = a \end{align*} as \displaystyle \displaystyle \begin{align*} A = \int_a^7{ \sqrt{ 49 - x^2 } \, dx } \end{align*}.

Equating this to the known area we have \displaystyle \displaystyle \begin{align*} \int_a^7{ \sqrt{ 49 - x^2 } \, dx } &= \frac{ 49 \, \pi }{ 6} \end{align*}.

In order to evaluate this integral and solve for the value of a, we need to make a trigonometric substitution \displaystyle \displaystyle \begin{align*} x = 7\sin{(\theta)} \implies dx = 7\cos{(\theta)}\,d\theta \end{align*} and note that when \displaystyle \displaystyle \begin{align*} x = a , \theta = \arcsin{ \left( \frac{a}{7} \right) } \end{align*} and when \displaystyle \displaystyle \begin{align*} x = 7 , \theta = \frac{\pi}{2} \end{align*}. Substituting gives

\displaystyle \displaystyle \begin{align*} \int_{\arcsin{ \left( \frac{a}{7} \right) } } ^{ \frac{\pi}{2} } { \sqrt{ 49 - \left[ 7\sin{(\theta)} \right] ^2 } \cdot 7\cos{(\theta)}\,d\theta } &= \frac{49\,\pi}{6} \\ 7 \int_{\arcsin{ \left( \frac{a}{7} \right) } }^{\frac{\pi}{2}} { \sqrt{ 49 \left[ 1 - \sin^2{ ( \theta ) } \right] } \cos{ ( \theta) } \, d\theta } &= \frac{49 \, \pi }{6} \\ 7 \int_{ \arcsin{ \left( \frac{a}{7} \right) } }^{ \frac{ \pi }{ 2 } }{ 7\cos^2{(\theta)}\,d\theta } &= \frac{ 49 \, \pi}{6} \\ \frac{ 49 }{2} \int_{ \arcsin{ \left( \frac{a}{7} \right) } }^{ \frac{ \pi }{2}}{ 1 + \cos{ (2\theta)} \,d\theta } &= \frac{ 49 \, \pi }{6} \\ \int_{\arcsin{\left( \frac{a}{7} \right) }}^{\frac{\pi}{2}}{1 + \cos{(2\theta)}\,d\theta} &= \frac{\pi}{3} \\ \left[ \theta + \frac{1}{2}\sin{(2\theta)} \right] _{\arcsin{\left( \frac{a}{7} \right) } }^{\frac{\pi}{2}} &= \frac{\pi}{3} \end{align*}

\displaystyle \displaystyle \begin{align*} \left[ \frac{\pi}{2} + \frac{1}{2} \sin{ \left( 2 \cdot \frac{\pi}{2} \right) } \right] - \left\{ \arcsin{ \left( \frac{a}{7} \right) } + \frac{1}{2} \sin{ \left[ 2\arcsin{ \left( \frac{a}{7} \right) } \right] } \right\} &= \frac{\pi}{3} \\ \frac{\pi}{2} - \arcsin{ \left( \frac{a}{7} \right) } - \sin{ \left[ \arcsin{ \left( \frac{a}{7} \right) } \right] } \,\sqrt{ 1 - \sin^2{ \left[ \arcsin{ \left( \frac{a}{7} \right) } \right] } } &= \frac{\pi}{3} \\ -\arcsin{ \left( \frac{a}{7} \right) } - \frac{a}{7} \, \sqrt{ 1 - \left( \frac{a}{7} \right) ^2 } &= \frac{\pi}{3} - \frac{\pi}{2} \\ -\arcsin{ \left( \frac{a}{7} \right) } - \frac{a}{7} \, \sqrt{ \frac{49 - a^2}{49} } &= -\frac{\pi}{6} \\ \arcsin{ \left( \frac{a}{7} \right) } + \frac{ a \, \sqrt{ 49 - a^2 }}{49} &= \frac{\pi}{6} \end{align*}

Unfortunately there is no way to evaluate the value of a here exactly, but you should be able to get a numerical answer using a CAS :)
• May 19th 2013, 11:06 PM
pandakrap
Re: Drawing vertical line through a circle to find for what x does that vertical line
:) Brilliant! Thank you so much.
• May 20th 2013, 01:57 AM
BobP
Re: Drawing vertical line through a circle to find for what x does that vertical line
If the ends of a chord drawn in a circle radius $\displaystyle r$ subtend an angle $\displaystyle \theta$ at the centre of the circle, then the area 'cut off' by the chord is $\displaystyle r^{2}(\theta - \sin \theta)/2.$

That means we need $\displaystyle r^{2}(\theta - \sin \theta)/2=\pi r^{2}/3,$ or, $\displaystyle \theta - \sin \theta=2\pi /3.$

That requires a numerical solution, which turns out to be $\displaystyle \theta = 2.60533.$

For a circle of radius $\displaystyle 7,$ it follows that the centre of the chord should be at a distance of $\displaystyle 1.8545$ (approx) from the centre of the circle.
• May 20th 2013, 02:23 AM
ibdutt
Re: Drawing vertical line through a circle to find for what x does that vertical line
Very well done in fact i was stuck at the same stage. i used the formula ∫▒√(a^2- x^2 ) dx= x/2 √(a^2- x^2 )+ 1/2 sin^(-1)⁡〖x/a〗+c directly the issue remains how to get a ??
• May 20th 2013, 02:26 AM
Prove It
Re: Drawing vertical line through a circle to find for what x does that vertical line
You need to use a numerical method and/or technology such as a CAS. There is no way to isolate a exactly when it's both inside and outside of a transcendental function (unless it's a very special case where a solution sticks out at you).