لو سمحتم احد يساعدني

26-11-2009, 04:02 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, 10:11 PM

أين معادلة الدائرة ؟

28-11-2009, 10:32 AM

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

^ هي الاس

ومشكورين

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

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

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

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

$(y-3)^2 = 12$

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

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

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

28-11-2009, 08:08 PM

مشكور اخوي

بس باقي الفقراات

ياليت تساعدني في الفقره بي وتشرحها لنا اذا ماعليك امر

30-11-2009, 02:08 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, 02:24 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, 02:52 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, 05:37 PM

يعطيك الف عافيه اخووي

والله يوفقك يارب

دنيا واخره


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