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- Jan 26, 2012

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- Thread starter
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- #1

- Jan 26, 2012

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- Jan 26, 2012

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1) anemone

2) MarkFL

3) eddybob123

4) Opalg

Solution (from Opalg):

In the diagram, the centre $O$ of the circle is three-quarters of the way along the diagonal from $C$ to $A$, and so $OC = \frac34\sqrt2$. The radius $OT$ is $\frac14$, and by Pythagoras, $TC^2 = \frac{18}{16} - \frac1{16}$, hence $TC = \frac14\sqrt{17}.$ Therefore $\tan\alpha = \dfrac1{\sqrt{17}}$, and $\tan\beta = \tan(45^\circ - \alpha) = \dfrac{1-\frac1{\sqrt{17}}}{1+\frac1{\sqrt{17}}} = \dfrac{\sqrt{17}-1}{\sqrt{17}+1} = \dfrac{9-\sqrt{17}}8$ (the last equality comes from multiplying top and bottom of the fraction by ${\sqrt{17}-1}$). Then $ED = \tan\beta$, and the shaded area is $\frac12ED*CD = \dfrac{9-\sqrt{17}}{16} \approx 0.3048.$

Alternate solution (from MarkFL):

\(\displaystyle y-1=m(x-1)\)

(1) \(\displaystyle y=mx+1-m\)

Suppose we wish to generalize a bit and let the radius of the circle be \(\displaystyle 0<r<\frac{1}{2}\).

Next, to determine $m$, we may use the formula for the distance between a point and a line. A derivation of this formula is given here:

http://mathhelpboards.com/math-notes-49/finding-distance-between-point-line-2952.html

The point, the center of the circle, is $(r,r)$, and the line is given in (1), and so we may state:

\(\displaystyle r=\frac{|mr+1-m-r|}{\sqrt{m^2+1}}\)

\(\displaystyle r\sqrt{m^2+1}=|(m-1)(r-1)|\)

Squaring both sides, we obtain after simplification:

\(\displaystyle (1-2r)m^2-2(r-1)^2m+(1-2r)=0\)

Applying the quadratic formula, we find:

\(\displaystyle m=\frac{(r-1)^2\pm r\sqrt{r^2-4r+2}}{1-2r}\)

Observing that we want the greater of the two roots as the smaller root represents the tangent line that passes over the circle, we are left with:

\(\displaystyle m=\frac{(r-1)^2+r\sqrt{r^2-4r+2}}{1-2r}\)

Now, to find the base $b$ of the triangular area, we may write:

\(\displaystyle m=\frac{(r-1)^2+r\sqrt{r^2-4r+2}}{1-2r}=\frac{1}{b}\implies b=\frac{(r-1)^2-r\sqrt{r^2-4r+2}}{1-2r}\)

Hence, we find the shaded area $A$ to be:

\(\displaystyle A(r)=\frac{1}{2}bh=\frac{1}{2}\left(\frac{(r-1)^2-r\sqrt{r^2-4r+2}}{1-2r} \right)(1)=\frac{(r-1)^2-r\sqrt{r^2-4r+2}}{2(1-2r)}\)

Now, for the given problem, we have \(\displaystyle r=\frac{1}{4}\), and so the value of $A$ for this particular value of $r$ is:

\(\displaystyle A\left(\frac{1}{4} \right)=\frac{\left(\frac{1}{4}-1 \right)^2-\frac{1}{4}\sqrt{\left(\frac{1}{4} \right)^2-4\left(\frac{1}{4} \right)+2}}{2\left(1-2\left(\frac{1}{4} \right) \right)}=\frac{9-\sqrt{17}}{16}\)

I'm also including anemone's solution because of how thorough it is:

Also, by definition, the gradient of the straight line is also the tangent of the angle that the line makes with the horizontal, it's not hard to deduce that the gradient of the normal line (the green line as shown in the diagram) is therefore represented by the value $-\tan \theta$ and this gives the gradient of the tangent line as the negative reciprocal of $-\tan \theta$, i.e. \(\displaystyle \frac{1}{\tan \theta}\).

Since the normal line passes through the center of the circle, its equation can then be formed by using the formula $y-y_1=m(x-x_1)$ and in this case, the equation of the normal line is given by

\(\displaystyle y-\frac{1}{4}=-\tan \theta(x-\frac{1}{4})\)

and the equation of the tangent is given by

\(\displaystyle y-1=\frac{1}{\tan \theta}(x-1)\)

We also use the fact that this circle can be re-defined as the locus of all points that satisfy the equations \(\displaystyle x=a+r\cos t\) and \(\displaystyle y=b+r\sin t\) where $x$ and $y$ are the coordinates of any point on the circle and $r$ is the radius of the circle and $t$ is the parameter, the angle subtended by the point a the circle's center and $(a, b)$ are the center of the circle.

Thus, the intersection point between the tangent to the circle and the circle is \(\displaystyle \left(\frac{1}{4}+\frac{\cos \theta}{4},\;\frac{1}{4}-\frac{\sin \theta}{4}\right)\).

Now to solve for the value for $\tan \theta$, what we need to do is by substituting the coordinates of the intersection point between the curve/the normal line and the tangent line, \(\displaystyle \left(\frac{1}{4}+\frac{\cos \theta}{4},\;\frac{1}{4}-\frac{\sin \theta}{4}\right)\) back into either the equation of the normal or tangent line, and get

\(\displaystyle y-1=\frac{1}{\tan \theta}(x-1)\)

\(\displaystyle \frac{1}{4}-\frac{\sin \theta}{4}-1=\frac{\cos \theta}{\sin \theta}(\frac{1}{4}+\frac{\cos \theta}{4}-1)\)

\(\displaystyle -\frac{3}{4}-\frac{\sin \theta}{4}=\frac{\cos \theta}{\sin \theta}(-\frac{3}{4}+\frac{\cos \theta}{4})\)

\(\displaystyle -3-\sin \theta=\frac{\cos \theta}{\sin \theta}(-3+\cos \theta)\)

\(\displaystyle -3\sin \theta-\sin^2 \theta=-3\cos \theta +\cos^2 \theta\)

\(\displaystyle 3\cos \theta-3\sin \theta=\cos^2 \theta+\sin^2 \theta=1\)

\(\displaystyle (3\cos \theta-3\sin \theta)^2=1\)

\(\displaystyle 9-9\sin\ 2 \theta=1\)

\(\displaystyle \sin\ 2 \theta=\frac{8}{9}\)

and hence \(\displaystyle \tan\ 2 \theta=\frac{8}{\sqrt{17}}\)

and by using the double angle formula for $\tan 2 \theta$, we thus obtain

\(\displaystyle \frac{2\tan \theta}{1-\tan^2 \theta}=\frac{8}{\sqrt{17}}\)

and solving this quadratic formula for $\tan \theta$, we ended up getting \(\displaystyle \tan \theta=\frac{9-\sqrt{17}}{8}\) since $\tan \theta >0$.

We now obtain the equation of the tangent line, \(\displaystyle y-1=\frac{1}{\left(\frac{9-\sqrt{17}}{8}\right)}(x-1)\) and when this tangent line intersects with the x-axis (i.e. $y=0$), its corresponding $x$ value is

\(\displaystyle 0-1=\frac{1}{\left(\frac{9-\sqrt{17}}{8}\right)}(x-1)\)

\(\displaystyle -\left(\frac{9-\sqrt{17}}{8}\right)=x-1\)

\(\displaystyle x=\frac{\sqrt{17}-1}{8}\)

Now, we have the base and the height of the right-angled triangle and

\(\displaystyle \text{the area of the shaded region}= \text{the area of this right-angled triangle}=\frac{1}{2}(1)(1-\frac{\sqrt{17}-1}{8})=\frac{9-\sqrt{17}}{16}\)

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