State the center and radius..Given the equation of a circle, (x+2)^2+(y-1)^2=36, state the center and radius.
The standard equation of a circle with center ( a , b) and a radius of r is given as: (x - a)^2 + (y-b)^2 = r^2
Now we have the equation of the circle of (x+2)^2+(y-1)^2=36
Comparing this with the standard form (x-a)^2 + (y-b)^2 = r^2, we see
a = -2 , b = 1 , r = 6
Therefore the radius is 6 and the center is (-2 , 1).
The equation of circle is given the (x-h)^2+(y-h)^2 = r^2...(1), where (h,k) is the coordinates of the centre. and r is the radius of the circle.
The given equation is (x+2)^2 +(y-1)^2 = 36....(2)
If (1) and (2) are representing the same circle, then the like terms , we get :
(x-h)^2 = (x+2)^2.
-2xh = 4x and +h^2 = 2^2 . Therefore h = 4x/-2x = -2.
Similarly, ((y-k)^2 = (y-1)^2 gives: k= 1.
r^2 = 36 , r = (36)^(1/2) = 6.
Therefore the centre (h, k) = (-2,1).
radius r = 6.
So the centre of the circle (x+2)^2+(y-1) = 36 is at (-2,1). The radius is 6.
We'll write the equation of the circle:
C(h , k) - the x and y coordinates of the center of the circle
r - the radius of the circle
Comparing the given equation and the equation of the circle, with perfect squares, we'll identify the followings:
h = -2
k = 1
r^2 = 36
r = 6
The circle has the coordinates of the center C(-2 , 1) and the radius r = 6.