+0  
 
0
449
2
avatar+206 

Find the area between two concentric circles defined by 

x2 + y2 -2x + 4y + 1 = 0 

x2 + y2 -2x + 4y - 11 = 0 

LunarRoxey  Apr 5, 2017
 #1
avatar+87301 
+3

Let's put these into standard form, first

 

x^2 + y^2 -2x + 4y + 1 = 0 

x^2 - 2x + y^2 + 2x =  -1       complete the square on x and y

x^2 - 2x + 1 + y^2 + 2x + 4  =  -1 + 1 + 4     factor

(x - 1)^2 + ( y + 2)^2  =  4

This is a circle centered at (1, -2) with a radius of 2

 

x^2 + y^2 -2x + 4y - 11 = 0

x^2 - 2x + y^2+ 4y = 11

x^2 - 2x + 1 + y^2 + 4y + 4  = 11 + 1 + 4

(x - 1)^2  + (y + 2)^2  = 16

This is  a circle with the same center and a radius of 4

 

The area between the concentric circles =

 

pi [ 4^2 - 2^2]   = pi [16 - 4 ]  =  12pi units^2  ≈  37.7 units^2

 

 

cool cool cool

CPhill  Apr 5, 2017
 #2
avatar+19632 
+3

Find the area between two concentric circles defined by 

 

Let xc the center of the circles in x

Let yc the center of the circles in y

 

\(x2 + y2 -2x + 4y \underbrace{+1}_{=x_c^2+y_c^2-r_1^2} = 0 \\\\ x2 + y2 -2x + 4y \underbrace{-11}_{=x_c^2+y_c^2-r_2^2} = 0 \)

 

\(\begin{array}{|lrcll|} \hline (1) & 1 &=& x_c^2+y_c^2-r_1^2 \\ (2) & -11 &=& x_c^2+y_c^2-r_2^2 \\ \hline (1)-(2): & 1-(-11) &=& x_c^2+y_c^2-r_1^2-(x_c^2+y_c^2-r_2^2) \\ & 1+11 &=& x_c^2+y_c^2-r_1^2-x_c^2-y_c^2+r_2^2 \\ & 12 &=& -r_1^2 +r_2^2 \\ & \mathbf{r_2^2-r_1^2} & \mathbf{=} & \mathbf{12} \\ \hline \end{array}\)

 

The area between two concentric circles:

\(\begin{array}{|rcll|} \hline A &=& \pi r_2^2 - \pi r_1^2 \\ &=& \pi \cdot ( r_2^2 - r_1^2) \quad & | \quad r_2^2-r_1^2 = 12 \\ &=& \pi \cdot 12 \\ &=& 37.6991118431 \\ \hline \end{array} \)

 

 

laugh

heureka  Apr 6, 2017

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.