لو سمحتم احد يساعدني

26-11-2009, 05:02 PM

Consider the circle C1 given by the equation .
(a) (4 marks) : Find the radius r and center M of C1 and determine the points A and B of intersection of C1 with the y-axis.

(b)(4 marks):Find the point D of intersection of the tangent lines to the circle at A
and B.

(c) (4 marks): Find the equation of the circle C2 with MD as a diameter.

(d) (3 marks): Obtain the parametric equations of the circle C1 .

مشكورين مقدما

26-11-2009, 11:11 PM

أين معادلة الدائرة ؟

28-11-2009, 11:32 AM

x^2+y^2+4x-6y-3=0

^ هي الاس

ومشكورين

28-11-2009, 02:53 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

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

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

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

$(y-3)^2 = 12$

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

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

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

28-11-2009, 09:08 PM

مشكور اخوي

بس باقي الفقراات

ياليت تساعدني في الفقره بي وتشرحها لنا اذا ماعليك امر

30-11-2009, 03:08 PM

Equation of the tangent T₁ at A :

$y - y_{0} = m(x-x_{0})$

where m is the slope and x₀ ,y₀ are the coordinates of A

Now , MA is perpendicular to T₁

$m_{MA}\times m_{T_{1}} = - 1$

$m_{MA}= \frac{y_{A}-y_{M}}{x_{A}-x_{M}}= \frac{3-2\sqrt{3}-3}{0-(-2)}=-\sqrt{3}$

$m_{T_{1}}=\frac{-1}{-\sqrt{3}}= \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

Equation of T₁

$y - y_{0} = m(x-x_{0})$

$y -(3-2\sqrt{3})= \frac{\sqrt{3}}{3}(x-0)\Rightarrow y=\frac{\sqrt{3}}{3}x + 3 - 2\sqrt{3}$

Same method for T₂ , the tangent at B

$y=\sqrt{3}x + 3 + 2\sqrt{3}$

For the point of intersection D , set y = y

$\frac{\sqrt{3}}{3}x + 3 - 2\sqrt{3}=\sqrt{3}x + 3 + 2\sqrt{3}$

Solving : x = -6

$y = 3 -4\sqrt{3}$

$D(-6,3 -4\sqrt{3})$

30-11-2009, 03:24 PM

(c) (4 marks): Find the equation of the circle C2 with MD as a diameter.

The center of C₂ is the midpoint of MD , call it N

$M(-2,3) , D(-6,3-4\sqrt{3})\Rightarrow N(-4,3-2\sqrt{3})$

The radius is NM

$NM=\sqrt{(x_{M}-x_{N})^2 +(y_{M}-y_{N})^2}=\sqrt{(-2+4)^2 +(3-3+2\sqrt{3})^2}=\sqrt{4+12}= 4$

The equation of the circle

$(x-x_{N})^2 +(y-y_{N})^2= r^2\Rightarrow (x+4)^2 + (y-3+2\sqrt{3})^2 = 16$

30-11-2009, 03:52 PM

(d) (3 marks): Obtain the parametric equations of the circle C1 .

A circle with centre (a,b) and radius r has parametric equations
x=a+ r cosθ, y = b+ r sinθ.

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

x=-2 + 4 cosθ, y = 3+ 4 sinθ

30-11-2009, 06:37 PM

يعطيك الف عافيه اخووي

والله يوفقك يارب

دنيا واخره


العضو المميز	الموضوع المميز	المشرف المميز
المنتدى متاح للتصفح فقط ولا يقبل المشاركات الجديدة	ماشاء الله تبارك الله ماشاء الله لاقوة الا بالله	المنتدى متاح للتصفح فقط ولا يقبل المشاركات الجديدة