SATHEE: Circle 1 Question 19

Circle 1 Question 19

19. A circle passes through three points $A, B$ and $C$ with the line segment $A C$ as its diameter. $A$ line passing through $A$ intersects the chord $B C$ at a point $D$ inside the circle. If angles $D A B$ and $C A B$ are $α$ and $β$ respectively and the distance between the point $A$ and the mid-point of the line segment $D C$ is $d$ , prove that the area of the circle is

$\frac{π d^{2} \cos^{2} α}{\cos^{2} α + \cos^{2} β + 2 \cos α \cos β \cos (β - α)}$

$(1996, 5 M)$

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Solution:

Let the radius of the circle be $r$ . Take $X$ -axis along $A C$ and the $O (0, 0)$ as centre of the circle. Therefore, coordinate of $A$ and $C$ are $(- r, 0)$ and $(r, 0)$ , respectively.

Now, $∠ B A C = β, ∠ B O C = 2 β$

Therefore, coordinates of $B$ are $(r \cos 2 β, r \sin 2 β)$ .

And slope of $A D$ is $\tan (β - α)$ .

Let $(x, y)$ be the coordinates of the point $D$ . Equation of $A D$ is

$y = \tan (β - α) (x + r)$

$[∵$ slope $= \tan (β - α)$ and point is $(- r, 0)]$

Now, equation of $B C$ is

$\begin{aligned} y = \frac{r \sin 2 β - 0}{r \cos 2 β - r} (x - r) \\ \Rightarrow & y & = \frac{r \cdot 2 \sin β \cos β}{r (- 2 \sin^{2} β)} (x - r) \\ \Rightarrow & y & = \frac{2 \sin β \cos β}{- 2 \sin^{2} β} (x - r) \\ \Rightarrow & y & = - \cot β (x - r) \end{aligned}$

To obtain the coordinate of $D$ , solve Eqs. (i) and (ii) simultaneously

$\begin{array}{lc} \Rightarrow & \tan (β - α) (x + r) = - \cot β (x - r) \\ \Rightarrow & x \tan (β - α) + r \tan (β - α) = - x \cot β + r \cot β \\ \Rightarrow x [\tan (β - α) + \cot β] = r [\cot β - \tan (β - α)] \\ \Rightarrow x \frac{\sin (β - α)}{\cos (β - α)} + \frac{\cos β}{\sin β} = r \frac{\cos β}{\sin β} - \frac{\sin (β - α)}{\cos (β - α)} \end{array}$

$\begin{matrix} \Rightarrow x \frac{\sin (β - α) \sin β + \cos (β - α) \cos β}{\cos (β - α) \sin β} \\ = r \frac{\cos β \cos (β - α) - \sin β \sin (β - α)}{\sin β \cos (β - α)} \\ \Rightarrow x [\cos (β - α - β)] = r [\cos (β - α + β)] \\ \Rightarrow x = \frac{r \cos (2 β - α)}{\cos α} \end{matrix}$

On putting this value in Eq. (ii), we get

$\begin{aligned} y & = - \cot β \frac{r \cos (2 β - α)}{\cos α} - r \\ \Rightarrow y & = - \frac{\cos β \cdot r}{\sin β} \frac{\cos (2 β - α) - \cos α}{\cos α} \\ \Rightarrow y & = - \frac{r \cos β}{\sin β} \frac{2 \sin \frac{2 β - α + α}{2} \sin \frac{α - 2 β + α}{2}}{\cos α} \\ \Rightarrow y & = - \frac{r \cos β}{\sin β} \frac{2 \sin β \cdot \sin (α - β)}{\cos α} \\ = - 2 r \cos β \sin (α - β) / \cos α \end{aligned}$

Therefore, coordinates of $D$ are

$\frac{r \cos (2 β - α)}{\cos α}, - \frac{2 r \cos β \sin (α - β)}{\cos α}$

Thus, coordinates of $E$ are

$\begin{aligned} \frac{r \cos (2 β - α) + r \cos α}{2 \cos α}, - r \frac{\cos β \sin (α - β)}{\cos α} \\ \Rightarrow r \frac{2 \cos \frac{2 β - α + α}{2} \cdot \cos \frac{2 β - α - α}{2}}{2 \cos α}, \\ \Rightarrow r \frac{\cos β \cdot \cos (β - α)}{\cos α}, r \frac{\cos β \sin (β - α)}{\cos α} \end{aligned}$

Since, $A E = d$ , we get

$\begin{aligned} d^{2} = r^{2} \frac{\cos β \cos (β - α)}{\cos α} + 1^{2} + r^{2} \frac{\cos β \sin (β - α)}{\cos α}^{2} \\ = \frac{r^{2}}{\cos^{2} α} [\cos^{2} β \cos^{2} (β - α) + \cos^{2} α \\ + 2 \cos β \cos (β - α) \cos α + \cos^{2} β \sin^{2} (β - α)] \\ = \frac{r^{2}}{\cos^{2} α} [\cos^{2} β \cos^{2} (β - α) + \sin^{2} (β - α) + \cos^{2} α \\ + 2 \cos β \cos α \cos (β - α)] \\ = \frac{r^{2}}{\cos^{2} α} [\cos^{2} β + \cos^{2} α + 2 \cos α \cos β \cos (β - α)] \\ \Rightarrow r^{2} = \frac{d^{2} \cos^{2} α}{\cos^{2} β + \cos^{2} α + 2 \cos α \cos β \cos (β - α)} \end{aligned}$