Answer:
![\huge\boxed{Center = (3,3),Radius = 2\sqrt{15} }](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7BCenter%20%3D%20%283%2C3%29%2CRadius%20%3D%202%5Csqrt%7B15%7D%20%7D)
Step-by-step explanation:
<u><em>Given equation is</em></u>
![x^2 + y^2 -6x-6y -42 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20-6x-6y%20-42%20%3D%200)
Adding 42 to both sides
![x^2 + y^2 -6x-6y = 42\\](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20-6x-6y%20%3D%2042%5C%5C)
Completing squares
![x^2 -6x+y^2 -6x = 42\\(x)^2 - 2(x)(3) +y^2 - 2(y)(3) = 42](https://tex.z-dn.net/?f=x%5E2%20-6x%2By%5E2%20-6x%20%3D%2042%5C%5C%28x%29%5E2%20-%202%28x%29%283%29%20%2By%5E2%20-%202%28y%29%283%29%20%3D%2042)
Adding (3)² => 9 and (3)² => 9 to both sides
![(x-3)^2+(y-3)^2 = 42+9+9\\(x-3)^2 + (y-3)^2 = 60\\(x-3)^2 (y-3)^2 = (2{\sqrt{15})^2}](https://tex.z-dn.net/?f=%28x-3%29%5E2%2B%28y-3%29%5E2%20%3D%2042%2B9%2B9%5C%5C%28x-3%29%5E2%20%2B%20%28y-3%29%5E2%20%3D%2060%5C%5C%28x-3%29%5E2%20%20%28y-3%29%5E2%20%3D%20%282%7B%5Csqrt%7B15%7D%29%5E2%7D)
Comparing it with
where Center = (h,k) and Radius = r
We get:
Center = (3,3)
Radius = ![2\sqrt{15}](https://tex.z-dn.net/?f=2%5Csqrt%7B15%7D)
The easiest way is to try the point (-4,1), that is, x=-4, y=1,
to see which equation works.
b works.
The usual way to do it is to find the equation of the circle
standard form of a circle is (x-h)²+(y-k)²=r², (h,k) are the coordinates of the center, r is the radius.
in this case, the center is (-2,1), so (x+2)²+(y-1)²=r²
the given point (-4,1) is for you to find r: (-4+2)²+(1-1)²=r², r=2
so the equation is (x+2)²+(y-1)²=2²
expand it: x²+4x+4+y²-2y+1=4
x²+y²+4x-2y+1=0, which is answer b.
Answer:
distributive property
Step-by-step explanation:
The distributive property is described as
a(b + c) = ab + ac
5(12 - 4) = 5(12) - 5(4) ← is obtained using the distributive property
Answer: 5
Step-by-step explanation: