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 3 . Then the square of the length of the chord of the circle on the line x = 1 is _______.
A. A
The equation of the circle is x^2 + y^2 + 2gx + 2fy + 25 = 0. The centre of the circle is (-g, -f). Since it lies in the first quadrant, -g > 0 and -f > 0, which implies g The centre lies on the line 2x - y = 4, so: -2g - (-f) = 4 f = 2g + 4 Let the radius of the circle be R. The side length a of an equilateral triangle inscribed in the circle is a = 3 R. The area of the equilateral triangle is given as 27 3 : 3 4 a^2 = 27 3 3 4 (3R^2) = 27 3 R^2 = 36 The radius of the circle is also given by R^2 = g^2 + f^2 - c. Here c = 25, so: g^2 + f^2 - 25 = 36 g^2 + f^2 = 61 Substituting f = 2g + 4 into the equation: g^2 + (2g + 4)^2 = 61 5g^2 + 16g - 45 = 0 (5g - 9)(g + 5) = 0 Since g Then, f = 2(-5) + 4 = -6. The centre of the circle is (5, 6) and its radius is R = 6. The distance d from the centre (5, 6) to the line x = 1 is: d = |5 - 1| = 4 The length of the chord L on the line x = 1 is given by: L = 2 R^2 - d^2 = 2 36 - 16 = 2 20 The square of the length of the chord is: L^2 = (2 20 )^2 = 4 20 = 80 Answer: 80
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