Answer:
![Equation : \\(x+3)^2 + ( y -2)^2 = 36](https://tex.z-dn.net/?f=Equation%20%3A%20%5C%5C%28x%2B3%29%5E2%20%2B%20%28%20y%20-2%29%5E2%20%3D%2036)
Step-by-step explanation:
For standard form the circle's equation we need the centre of the circle and the radius.
Step 1: <em><u>Find the centre</u></em>
If the centre is not given find the end points of the diameter
and then find the mid point.
Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )
The mid-point of the diameter is :
![Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) = (-3, 2)](https://tex.z-dn.net/?f=Mid-point%20%3D%20%28%5Cfrac%7B-3%20%2B%20-%203%7D%7B2%7D%2C%20%5Cfrac%7B-4%2B8%7D%7B2%7D%29%20%3D%20%20%28-3%2C%202%29)
Therefore, centre of the circle = ( -3 , 2 )
Step 2 : <u>Find radius</u>
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Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )
That is ,
![Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\](https://tex.z-dn.net/?f=Diameter%20%3D%20%5Csqrt%7B%28-3-%28-3%29%29%5E2%20%2B%20%28%20-4%20-8%29%5E2%7D%5C%5C)
![= \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12](https://tex.z-dn.net/?f=%3D%20%5Csqrt%7B%28-3%20%2B%203%29%5E2%20%2B%20%28-12%29%5E2%7D%5C%5C%5C%5C%3D%5Csqrt%7B0%20%2B%20144%7D%5C%5C%5C%5C%3D12)
Therefore ,
![Radius = \frac{12}{2} = 6](https://tex.z-dn.net/?f=Radius%20%3D%20%5Cfrac%7B12%7D%7B2%7D%20%3D%206)
Step 3 : <u>Equation of the circle</u>
Standard equation of the circle with centre ( h ,k )
and radius ,r is :
![(x - h)^2+(y -k)^2 = r^2](https://tex.z-dn.net/?f=%28x%20-%20h%29%5E2%2B%28y%20-k%29%5E2%20%3D%20r%5E2)
Therefore, the equation of the circle with centre ( -3, 2)
and radius = 6 is :
![(x - (-3))^2 + (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36](https://tex.z-dn.net/?f=%28x%20-%20%28-3%29%29%5E2%20%2B%20%20%28y%20-%202%29%5E2%20%3D%206%5E2%5C%5C%5C%5C%28x%20%2B%203%29%5E2%20%2B%20%28y%20-%202%29%5E2%20%3D%2036)