Find the center and the radius of the circle with the equation: x^2-2x+y^2+4y+1=0?

Center (h,k) is (1,-2) and radius r = 2

Step-by-step explanation:

We need to find the center and radius of the circle of the given equation:

We need to transform the above equation into standard form of circle

where(h,k) is the center of circle and r is radius of circle.

Solving the given equation:

Moving 1 to right side

Now making perfect square of x^2-2x and y^2+4y

Adding +1 and +4 on both sides of the equation

Now, x^2-2x+1 is equal to (x-1)^2 and y^2+4y+2 =(y+2)^2

Comparing with standard equation of circle:

h = 1 , k =-2 and r =2 because r^2 =4 then r=2

So, center (h,k) is (1,-2) and radius r = 2

C(0,0)

R= 6.48

Step-by-step explanation:

I just did it and got it right

(x + 3)² + (y+ 2)² = 1²

Step-by-step explanation:

The standard form for the equation of a circle with centre (a, b) and radius r is

(x - a)² + (y - b)² = r²

x² + 6x + y² + 4y + 12 =  0

Step 1. Subtract the constant from each side

x² + 6x +           y² + 4y        = -12

Step 2. Complete the squares for x and y

Square half the coefficients of x and y

(x² + 6x + 3²) + (y² + 4y + 2²) = -12 + 3² + 2²

Step 3. Write the lhs as squares of binomials

(x + 3)² + (y+ 2)² = -12 + 3² + 2²

Step 4. Convert the rhs to a square

(x + 3)² + (y+ 2)² = -12 + 9 + 4

(x + 3)² + (y+ 2)² = 1

(x + 3)² + (y+ 2)² = 1²

The graph below shows that this is the equation for a circle with

centre (-3, -2) and radius 1.

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Step-by-step explanation: i did it by doing nun

Center = ( 1,-2)

radius = 2

Center is (-3, -2) and radius is 1

Step-by-step explanation:

By comparing the 2 equations,

⇒ 2gx = 6x

⇒ 2g = 6

∴ g = 6/2 = 3

⇒ 2fy = 4y

⇒ 2f = 4

∴ f = 4/2 = 2

⇒ c = 12

Center = (-g, -f) = (-3, -2)

Radius = √g² + f² - c

           = √3² + 2² - 12 = √9 + 4 - 12

           = 1

Center:

(−1,1)

Radius:

4

ANSWER B

centre = (- 3, - 2) , radius = 1

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + 6x + y² + 4y + 12 = 0 ( subtract 12 from both sides )

x² + 6x + y² + 4y = - 12

Use the method of completing the square on the x and y terms

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(3)x + 9 + y² + 2(2)y + 4 = - 12 + 9 + 4

(x + 3)² + (y + 2)² = 1 ← in standard form

with centre = (- (- 3), - (- 2)) and r² = 1, that is

centre = (- 3, - 2) and radius = 1

centre = (2, 6) and radius = 3

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

To obtain this form complete the square on both the x and y terms

express the equation as

x² - 4x + y² - 12y = - 31

x² + 2(- 2)x + 4 + y² + 2(- 6)y + 36 = - 31 + 4 + 36

(x - 2)² + (y - 6)² = 9 ← in standard form

with centre = (2, 6) and r = = 3