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 __________.

Hint:

Find the radius using the area of the equilateral triangle, determine the centre using the line equation and radius formula, then calculate the distance from the centre to $x=1$ to find the chord length.

Answer 1

Correct Answer: 80

Alternatively, we can find the intersection points of the circle and the line to find the chord length.
The equation of the circle with centre $(5, 6)$ and radius $R=6$ is:
$$ (x - 5)^2 + (y - 6)^2 = 36 $$
To find the chord on the line $x = 1$, we substitute $x = 1$ into the circle's equation:
$$ (1 - 5)^2 + (y - 6)^2 = 36 $$
$$ (-4)^2 + (y - 6)^2 = 36 \Rightarrow 16 + (y - 6)^2 = 36 $$
$$ (y - 6)^2 = 20 \Rightarrow y - 6 = \pm \sqrt{20} = \pm 2\sqrt{5} $$
So the coordinates of the endpoints of the chord are $(1, 6 + 2\sqrt{5})$ and $(1, 6 - 2\sqrt{5})$.
The length $L$ of this chord is the distance between these two points:
$$ L = \sqrt{(1 - 1)^2 + ((6 + 2\sqrt{5}) - (6 - 2\sqrt{5}))^2} = \sqrt{0^2 + (4\sqrt{5})^2} = 4\sqrt{5} $$
The square of the length of the chord is $L^2 = (4\sqrt{5})^2 = 16 \times 5 = 80$.