How do you find the center and radius of the circle #9x^2 + 54x + 9y^2 − 6y + 55 = 0#?

1 Answer
Nallasivam V
Oct 23, 2016

center #(-3, 1/3)#

Radius #sqrt3#

Explanation:

Given -

#9x^2+54x+9y^2-6y+55=0#

Let us rewrite the equation in the standard form -

#9(x^2+6x)+9(y^2-2/3y)=-55#

#9(x^2+6x+9)+9(y^2-2/3y+1/9)=-55+81+1#

#9(x^2+6x+9)+9(y^2-2/3y+1/9)=27#

#(x^2+6x+9)+(y^2-2/3y+1/9)=27/9=3#

#(x+3)^2+(y-1/3)^2=3#

Now it is in the form of -

(x-h)^2+(y-k)^2=a^2

In this case #(h, k)# is the center and #a# is the radius

Apply this to our case

center #(-3, 1/3)#

Radius #sqrt3#

Look at the image

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