Answer:
A. ![(x+5)^2+(y-4)^2=16](https://tex.z-dn.net/?f=%28x%2B5%29%5E2%2B%28y-4%29%5E2%3D16)
B. ![(x-1)^2+(y-4)^2=2\\](https://tex.z-dn.net/?f=%28x-1%29%5E2%2B%28y-4%29%5E2%3D2%5C%5C)
C. ![$\left(x-3\right)^2+\left(y-5\right)^2=9$](https://tex.z-dn.net/?f=%24%5Cleft%28x-3%5Cright%29%5E2%2B%5Cleft%28y-5%5Cright%29%5E2%3D9%24)
Step-by-Step Explanation:
A. Tangent to x-axis, r=4, contains (-5,8)
The equation of a circle is this: ![(x-h)^2+(y-k)^2=r^2\\](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2%5C%5C)
Therefore, we know that our equation will be: ![(x-h)^2+(y-k)^2=4^2=16](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3D4%5E2%3D16)
Because the circle is tangent to the x-axis, the point (-5,8) will be the top of the circle.
Therefore, the center is (-5,4), where -5=h and 4=k.
The answer is ![(x+5)^2+(y-4)^2=16](https://tex.z-dn.net/?f=%28x%2B5%29%5E2%2B%28y-4%29%5E2%3D16)
B. Diameter endpoints (2,5) and (0,3)
To solve for the diameter, we can use this equation: ![d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2 }](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_%7B1%7D-x_%7B2%7D%29%5E2%2B%28y_%7B1%7D-y_%7B2%7D%29%5E2%20%20%20%20%7D)
![d=\sqrt{(2-0)^2+(5-3)^2 }\\\\d=\sqrt{2^2+2^2}=\sqrt{4+4} =\sqrt{8}= 2\sqrt{2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%282-0%29%5E2%2B%285-3%29%5E2%20%20%20%20%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B2%5E2%2B2%5E2%7D%3D%5Csqrt%7B4%2B4%7D%20%3D%5Csqrt%7B8%7D%3D%202%5Csqrt%7B2%7D)
Because the radius is half of the diameter, ![r=\sqrt{2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B2%7D)
Therefore, we know that our equation will be: ![(x-h)^2+(y-k)^2=(\sqrt{2}) ^2=2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3D%28%5Csqrt%7B2%7D%29%20%5E2%3D2)
Then, we can decipher that h=1, as the midpoint of 0 and 2 is 1, and k=4, as the midpoint of 3 and 5 is 4.
The answer is ![(x-1)^2+(y-4)^2=2](https://tex.z-dn.net/?f=%28x-1%29%5E2%2B%28y-4%29%5E2%3D2)
C. Center on x=3, tangent to y-axis at (0,5)
We can see that the center is (3,5). Therefore, the equation is ![(x-3)^2+(y-5)^2=r^2\\](https://tex.z-dn.net/?f=%28x-3%29%5E2%2B%28y-5%29%5E2%3Dr%5E2%5C%5C)
To solve for r, we can substitute the point (0,5) into the equation.
![(0-3)^2+(5-5)^2=r^2\\(-3)^2+(0)^2=r^2\\r^2=9](https://tex.z-dn.net/?f=%280-3%29%5E2%2B%285-5%29%5E2%3Dr%5E2%5C%5C%28-3%29%5E2%2B%280%29%5E2%3Dr%5E2%5C%5Cr%5E2%3D9)
The answer is ![(x-3)^2+(y-5)^2=9](https://tex.z-dn.net/?f=%28x-3%29%5E2%2B%28y-5%29%5E2%3D9)
<em>Hope this helped!</em>