Find the center and radius of a circle by transforming this equation into a standard form for a circle?

Question

Find the center and radius of a circle by transforming this equation into a standard form for a circle: x^2 +y^2-10y+8x+27=0

1) standard form of equation
2)Radius
3) Center

Can you show the steps as well...Im having trouble particularly with transforming it into standard form

Answer 1

The standard form of a circle is: (x-h)^2 + (y-k)^2 = r^2, where the center is equal to the point (h,k)



To convert it into standard form, you MUST know how to complete the square. Completing the square basically adds a number to both sides of the equation to make the quadratic a perfect square. You want a perfect square to put the circle in standard form.



Ex:



X^2 + 2x = 0



You take the square of HALF the middle term [2] and add it to both sides of the equation to make a perfect square



x^2 + 2x + 1 = 0 + 1



(x^2 + 2x + 1) is a perfect square now : (x+1)^2



so it becomes (x + 1)^2 = 1

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ACTUAL PROBLEM:



You need to complete the square TWICE, once for X, once for Y



Step 1: Put like terms together [X with X, Y with Y, ] and bring constant to other side



x^2 + 8x + y^2 -10y=-27



Step 2: Complete the square for both X and Y. Dont forget to add whatever you add to the left side to the right side [so they balance out]



x^2 + 8x + 16 + y^2 -10y + 25 = -27 + 16 + 25



(x+4)^2 + (y-5)^2 = 14



Standard form: (x-h)^2 + (y-k)^2 = r^2



radius = root 14



Center = (h,k) = (-4,5)

Answer 2

The standard form for a circle is (x-a)^2 + (y-b)^2 = c^2, where (a,b) is the centre of the circle and c is the radius



x^2 +y^2-10y+8x+27=0

x^2+8x+y^2-10y+27=0

In order to make this equation to the standard form, adding 41 on both sides



(x^2+8x+16)+(y^2-10y+25)+27 = 25+16



(x+4)^2 + (y-5)^2 = 14 = (Sq,.rt(14))^2 - Standard form



Centre is (-4, 5) and radius is Sq,.rt(14))

Answer 3

complete the squares for x and y to put it into standard form:

(x - h)^2 + (y - k)^2 = r^2, there (h,k) is the center and r is the radius



x^2 + 8x + y^2 - 10y = -27

(x^2 + 8x + ____) + (y^2 - 10y + ------) = -27 + ____ + -------

where ____ = (8/2)^2 and ---- = (-10/2)^2

(x^2 + 8x + 16) + (y^2 - 10y + 25) = -27 + 16 + 25

(x + 4)^2 + (y - 5)^2 = 14



center (-4 , 5); radius = sqrt(14)



note that x - -4 = x + 4