Chapter 6

Coordinate Geometry of the Circle

The equation of a circle in both of its forms, how to get from one to the other by completing the square, and the tangent property that turns most circle questions into straight-line questions.

หลักสูตรแกนกลาง / สสวท.: การวัดและเรขาคณิต (Measurement and Geometry)ตรงกับ ม.4 เรขาคณิตวิเคราะห์ (เพิ่มเติม) เรื่องวงกลม
Common Core (US): G-GPE, G-C
Cambridge: IGCSE 0606 — Topic 8: Coordinate geometry of the circle

6.1 The two forms of the equation

Centre-radius form, and the expanded form

(x - a)^{2} + (y - b)^{2} = r^{2} ⟷ x^{2} + y^{2} + 2 g x + 2 f y + c = 0

The first form shows you the circle; the second hides it. Completing the square in $x$ and in $y$ turns the second into the first.

ตัวอย่าง

Find the centre and radius of $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

วิธีทำ

1.Group: $(x^{2} - 6 x) + (y^{2} + 4 y) = 12$
2.Complete both squares: $(x - 3)^{2} - 9 + (y + 2)^{2} - 4 = 12$
3. $(x - 3)^{2} + (y + 2)^{2} = 25$

ตอบ Centre $(3, - 2)$ , radius $5$

ข้อควรระวัง

The radius is $5$ , not $25$ . The right-hand side is $r^{2}$ , and forgetting to take the root is the most common single error in this chapter — partly because $25$ is a believable radius.

เคล็ดลับ

The centre's coordinates have the OPPOSITE signs to the numbers in the brackets. $(x - 3)^{2}$ gives $x = 3$ and $(y + 2)^{2}$ gives $y = - 2$ . Saying the bracket out loud as " $x$ minus three" makes the sign obvious.

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