Answer:
The center of this circle is found at (4, -2) and the radius is 6.
Step-by-step explanation:
The equation for a circle in standard for is ( x – h)² + ( y - k)² = r². In this equation, (h, k) is the center of the circle and r is the radius.
From the given equation, ( x – 4)² + ( y + 2)² = 36, h is already subtracted from x, so h = 4, but the k is not subtracted, so (y + 2) = [y - (-2)], and we see that k is -2. Therefore, the center of the circle is (4, -2) and 36 = r². By taking the square root of 36, we find that r = 6.
The center of this circle is found at (4, -2) and the radius is 6.
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Hope this will help.
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First of all we need to find a representation of C, so this is shown in the figure below.
So the integral we need to compute is this:
![I=\int_c 4dy-4dx](https://tex.z-dn.net/?f=I%3D%5Cint_c%204dy-4dx)
So, as shown in the figure, C = C1 + C2, so:
Computing first integral:![c_{1}: y-y_{0}=m(x-x_{0}) \rightarrow y=x](https://tex.z-dn.net/?f=c_%7B1%7D%3A%20y-y_%7B0%7D%3Dm%28x-x_%7B0%7D%29%20%5Crightarrow%20y%3Dx)
Applying derivative:
![dy=dx](https://tex.z-dn.net/?f=dy%3Ddx)
Substituting this value into
![I_{1}](https://tex.z-dn.net/?f=I_%7B1%7D)
Computing second integral:![c_{2}: y-y_{0}=m(x-x_{0}) \rightarrow y-0=-(x-8) \rightarrow y=-x+8](https://tex.z-dn.net/?f=c_%7B2%7D%3A%20y-y_%7B0%7D%3Dm%28x-x_%7B0%7D%29%20%5Crightarrow%20y-0%3D-%28x-8%29%20%5Crightarrow%20y%3D-x%2B8)
Applying derivative:
![dy=-dx](https://tex.z-dn.net/?f=dy%3D-dx)
Substituting this differential into
![I_{2}](https://tex.z-dn.net/?f=I_%7B2%7D)
![I_{2}=\int_{c_{2}} 4(-dx)-4dx=\int_{c_{2}} -8dx=-8\int_{c_{2}}dx](https://tex.z-dn.net/?f=I_%7B2%7D%3D%5Cint_%7Bc_%7B2%7D%7D%204%28-dx%29-4dx%3D%5Cint_%7Bc_%7B2%7D%7D%20-8dx%3D-8%5Cint_%7Bc_%7B2%7D%7Ddx)
We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:
![I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}](https://tex.z-dn.net/?f=I_%7B2%7D%3D%20-8%5Cint_%7B4%7D%5E%7B8%7D%7Ddx%3D-8%5Bx%5D%5Cright%7C_4%20%5E%7B8%7D%3D-8%288-4%29%20%5Crightarrow%20%5Cboxed%7BI_%7B2%7D%3D-32%7D)
Finally: