منتديات الرياضيات العربية - عرض مشاركة واحدة

28-11-2009, 01:53 PM

x² + y² +4x - 6y - 3 = 0

Center and readius

Method1

Completing the square

x² + 4x + y² - 6y - 3 = 0

x² + 4x + 4 - 4 + y² -6y + 9 - 9 - 3 = 0

$(x+2)^2 - 4 +(y-3)^2 -9 - 3 = 0$

$(x+2)^2 +(y-3)^2 = 16$

center M(-2 , 3) and radius r = 4

Method2

General Form :

x² + y² - 2ax - 2by + c = 0

then the center is M (a,b) and the radius r² = a²+ b² - c

In the equation

x² + y² +4x - 6y - 3 = 0

2a = 4 -
a = -2
2b = - 6 -
b = 3

(3, 2-) M

r² = a² + b² - c

$r^2 =(-2)^2 + 3^2 - (-3)$

r² = 16
r= 4

Intersection with y-axis

x = 0

x² + y² +4x - 6y - 3 = 0

y² - 6y - 3 = 0

$y=\frac{-b \pm\sqrt{b^2-4ac}}{2a}=\frac{6 \pm\sqrt{48}}{2}=\frac{6 \pm4\sqrt{3}}{2}= 3 \pm 2\sqrt{3}$

The points are

$A (0 , 3-2\sqrt{3}) , B(0 , 3+2\sqrt{3})$

Another Method

$(x+2)^2 +(y-3)^2 = 16$

x = 0

$(2)^2 +(y-3)^2 = 16$

$4 +(y-3)^2 = 16$

$(y-3)^2 = 16 - 4$

$(y-3)^2 = 12$

$y - 3 =\pm sqrt{12}$

$y - 3 =\pm 2sqrt{3}$

$y=3 \pm 2sqrt{3}$