AB is a chord of length 2ka of a circle of radius a. The tangents to the circle at A and B meet in C if k^7 is negligible calculate the area of ABC

Question

AB is a chord of length 2ka of a circle of radius a. The tangents to the circle at A and B meet in C. Show that, if k is so small compared with unity that $k^7$ is negligible, the area of the triangle ABC is $a^2k^3+\frac12 a^2k^5$.

This image is a rough idea of what I think is going on. (O is the center of the circle and b is the length of line AC and X is the intersection point of line OC and chord AB.

The area required can be given as $k\times a\times h$ I first noticed that getting b in terms of a and k was possible because Triangle OAC and Triangle OBC right-angled triangles so:$$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{(ka)^2}$$ $$b = \frac{ka}{1- k^2}$$

Using binomial expansion to $k^4$ I got $$b = ka(1 + \frac{1}{2}k^2 + \frac{3k^4}{8})$$

Using this I got h as follows: $$a^2 + b^2 = \sqrt{a^2-(ka)^2} +h$$ $$h = a^2 + [ka(1 + \frac{1}{2}k^2 + \frac{3k^4}{8})]^2 -a\sqrt{1-k}$$ Expanding $\sqrt{1-k}$ $$h = a^2 + [ka(1 + \frac{1}{2}k^2 + \frac{3k^4}{8})]^2-a[1-\frac{k}{2}-\frac{k^2}{8}-\frac{k^3}{16}-\frac{5k^4}{128} -\frac{7k^5}{256}]$$ Expanding the middle term ,ignoring powers of k higher than 5:

$$h = a^2 + k^2a^2+k^4a^2 + 1-\frac{ak}{2}-\frac{ak^2}{8}-\frac{ak^3}{16}-\frac{5ak^4}{128} -\frac{7ak^5}{256}$$ $$AREA = kah =ka^3 + k^3a^3+k^5a^3 + 1-\frac{a^2k^2}{2}-\frac{a^2k^3}{8}-\frac{a^2k^4}{16}-\frac{5a^2k^5}{128} -\frac{7a^2k^6}{256}$$

The other way I had thought of was expressing h using triangle XAC instead of OBC but I kept making mistakes so I opted for this instead. I simply need a hint or a general idea of any another simpler method of going about the questions regardless of any errors I may have made in my working above.

Answer 1

In right triangle $OXB$, Pythagoras theorem gives :

$$OX^2+(ka)^2=a^2 \iff OX=a \sqrt{1-k^2}\tag{1}$$

In right triangle $OBC$, let us use the "Geometric mean theorem" stating that the length of altitude $BX$ is the geometric mean of the length of segments it creates on the hypotenuse , i.e., $OX$ and $XC$ :

$$OX.XC=XB^2 \iff XC=\frac{(XB)^2}{OX}=\frac{k^2a^2}{a \sqrt{1-k^2}}=\frac{k^2a}{\sqrt{1-k^2}}$$

As the area $\Delta$ of triangle $ABC$ is $XB.XC$, it remains to expand :

$$\Delta=ka.\frac{k^2a}{\sqrt{1-k^2}}=k^3a^2 (1-k^2)^{-1/2}$$

into a binomial series (as you have done) :

$$\Delta=k^3a^2(1+\tfrac12 k^2+...)$$

where the first omitted term is negligible.

Answer 2

$\square OACB$ is a kate with $\angle A=\angle B =90^\circ$ and $OA=OB=a.$ Since $\angle O=2k$, $AC=BC=a\tan k$ and $\angle C=\pi-2k$.

The area wanted is then $(ACB)=\frac12ABAC\sin\angle C=a^2\tan^2k\sin2k=a^2\sin^3k\sec k=a^2k^3+O(k^7).$