Question
Let the centre of the circle $x^{2} + y^{2} + 2gx + 2fy + 25 = 0$ be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is
$\_\_\_\_$ .
Solution
Sol. [Key Concept: The centre $( - g, - f)$ lies on $2x - y = 4$; the inscribed equilateral triangle area $\frac{3\sqrt{3}}{4}r^{2} = 27\sqrt{3}$ gives $r^{2} = 36$; the chord$2$ on $x = 1$ is $4(r^{2} - d^{2})$.]
$$x^{2} + y^{2} + 2gx + 2fy + 25 = 0 $$$C \equiv ( - g, - f)$ lies on line
$$\begin{array}{r} \therefore - 2\text{ }g + f = 4\#(i) \end{array}$$
Also $g^{2} + f^{2} - 25 = r^{2}$
Area of equation triangle inside the circle
$${= \frac{3\sqrt{3}}{4}r^{2} }{27\sqrt{3} = \frac{3\sqrt{3}}{4} \times r^{2} }{r^{2} = 36}$$
Solving (i) and (ii) $g = - 5,f = - 6$
$$\begin{matrix} & \mathcal{l}_{AB} = \sqrt{36 - 16} = 2\sqrt{20} \\ & \left( \mathcal{l}_{AB} \right)^{2} = 80 \end{matrix}$$
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