A-maths!!

x² + y² - 10x + 6y + 9 = 0

Since the circle cuts the x-axis at point A and point B.

Sub y = 0 into the equation, we get:

x² - 10x + 9 = 0
(x - 9)(x - 1) = 0
x = 1 or 9

So, the coordinates of A and B are (1,0) and (9,0).


Since the circle touches the y-axis at point C.

Sub x = 0 into the equation, we get:

y² + 6y + 9 = 0
(y + 3)² = 0
y = -3

So, the coordinates of C is (0,-3).


So, area of ΔABC
= 1/2 X length of AB X magnitude of y-coordinate of C
= 1/2 X (9-1) X 3
= 12 sq. units.

三角形底為 AB
高為 OC, O 係坐標中點

y = 0, x^2 -10x + 9 = 0
x = 9 or x = 1
so, AB = 8

x = 0, y^2 + 6y + 9 = 0
(y + 3)^2 = 0
y = -3
so, OC = 3

面積 = 1/2(8)(3) = 12

劃個圖出黎睇下就會明, 做呢 d 題目最重要劃圖,有圖令做法簡單好多