Phương trình đường tròn, cung và lượng giác trên đường tròn đơn vị

Hidden circle

• The SAT gives $x^2 + y^2 - 6x + 4y - 12 = 0$ and asks for the radius 半径 of the circle.
• The equation hides the centre and radius; completing the square reveals them.

Circle equations

• Standard form: $(x - h)^2 + (y - k)^2 = r^2$, centre $(h, k)$, radius $r$.
• Complete the square: $x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 4y = (y + 2)^2 - 4$. So $(x - 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25$: centre $(3, -2)$, radius $5$.

Completing the square gives a circle centred at (3, −2) with radius 5

Arcs and sectors 扇形

• A central angle $\theta$ (degrees) gives arc length 弧长 $\dfrac{\theta}{360} \times 2\pi r$ and sector area $\dfrac{\theta}{360} \times \pi r^2$.
• In radians: arc length $= r\theta$; $180° = \pi$ radians, so $60° = \dfrac{\pi}{3}$.

Tangents, inscribed angles, unit circle

• A tangent is perpendicular to the radius at the point of contact. An inscribed angle is half the central angle on the same arc.
• On the unit circle, the point at angle $\theta$ is $(\cos\theta, \sin\theta)$; $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, $\cos\dfrac{\pi}{3} = \dfrac{1}{2}$.

Traps

Recap