# Geometry problem on finding the diameter of a circle

$AB$ is the diameter of a circle. Tangents $AD$ and $BC$ are drawn so that $AC$ and $BD$ intersect at a point on the circle. If $|AD|=a$ and $|BC|=b$, $(a \neq b)$ then find the diameter of the circle.

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If the point on the circle, where the two "tangents" intersect, is important, give it a name ... –  Tony Piccolo Aug 9 '13 at 9:39
THE TWO TANGENTS ARE PARALLEL TANGENTS –  sangabattulalokesh Aug 9 '13 at 9:42
Sorry, I didn't read well the text. –  Tony Piccolo Aug 9 '13 at 9:46
Hint: 1) Mark the centre. 2) What is the angle between the intersecting lines? –  Parth Thakkar Aug 9 '13 at 9:50

$$\tan\theta = \frac{d}{a} = \frac{b}{d} \qquad \implies \qquad d^2 = a b \qquad \implies \qquad d = \sqrt{ab}$$

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No words to appreciate your way of solving . Just brilliant. –  Harish Kayarohanam Aug 9 '13 at 12:27
The simplest way I have ever seen –  sangabattulalokesh Aug 9 '13 at 16:03
It is a well known property of a right trapezium with perpendicular diagonals. –  Tony Piccolo Aug 9 '13 at 23:50

Consider the following diagram

$\hspace{3.2cm}$

Note that

$\triangle FBC$ is congruent to $\triangle FEC$
$\triangle FED$ is congruent to $\triangle FAD$

Thus, $\angle CFD$ is $\frac12$ of $\angle BFA$; therefore, $\angle CFD$ is a right angle.
Furthermore, $|DE|=|DA|=a$ and $|CE|=|CB|=b$

Since $\triangle CFD$ is a right triangle and $FE\perp CD$, we get that $\triangle CEF$ is similar to $\triangle FED$. This means that $$\frac{|CE|}{|EF|}=\frac{|EF|}{|ED|}\implies r^2=|EF|^2=|CE|\,|DE|=ab$$ Therefore, $r=\sqrt{ab}$ .

Oops, I misread the question. Here is an answer to the actual question:

$\hspace{3.2cm}$

$\triangle BAD$ is similar to $\triangle CBA$. Therefore, $$\frac{|BC|}{|AB|}=\frac{|AB|}{|AD|}\implies4r^2=|AB|^2=|AD||BC|=ab$$ Therefore, $r=\frac12\sqrt{ab}$.

A very similar answer, even though I misread the question the first time. I actually prefer the way I misread the question, so I will leave my original mis-answer.

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AC and BD should intersect at the circle, CD is not tangent to the circle. –  Greebo Aug 9 '13 at 11:40
@Greebo: Indeed. I've added a correction, but it is similar to the others. –  robjohn Aug 9 '13 at 15:58

Lets say AC and BD intersect at E , so at right angles as diameter subtends right angle at circumference . Using pythagorous theorem

$$\displaystyle\boxed{(diameter)^2 = AE^2 + BE^2}$$

So now we should calculate AE and BE to get d

as $$CE \times CA = CB^2 ---> 1$$ and $$DE \times DB = DA^2 ---> 2$$

and as the triangles AED and CEB are similar

$$\frac{AE}{EC} = \frac{DE}{EB} = \frac{a}{b}$$ $$\frac{AC}{EC} = \frac{a+b}{b} --------> 3$$ $$\frac{BD}{DE} = \frac{a+b}{a} --------> 4$$ substitute AC of result 3 in 1 and get $CE^2$ and from there CE , Similaryly get DE

$$CE = \frac{b*\sqrt{b}}{\sqrt{a+b}}$$

$$DE = \frac{a*\sqrt{a}}{\sqrt{a+b}}$$

so $$AE = \frac{a}{b} \times CE$$ $$\displaystyle\boxed{AE = \frac{a\sqrt{b}}{\sqrt{a+b}}}$$

similary

$$BE = \frac{b}{a} \times DE$$ $$\displaystyle\boxed{BE = \frac{b\sqrt{a}}{\sqrt{a+b}}}$$

so just substitute AE and BE in $(diameter)^2 = AE^2 + BE^2$and find d the diameter

$$d^2 = \frac{a^2*b}{a+b} + \frac{b^2*a}{a+b}$$ $$d^2 = \frac{ab*(a+b)}{a+b}$$

this implies

$$d = \sqrt{ab}$$

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