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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
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2+0 Answers

 #1
avatar+79821 
+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+18777 
+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

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