How do you find the center and radius for #x^2 + y^2 + 8y + 6 = 0#?

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Answer 1 · 2016-06-17T10:57:40

cetre at (0,-4)# and radius =#r = #sqrt10#.

The General Eqn. of a circle is given by #x^2+y^2+2gx+2fy+c=0,# provided that #g^2+f^2-c>0.#

Comparing the given eqn. with the general one, we get, #g=0,f=4,c=6, # so that, the condition #g^2+f^2-c=0+16-6=10>0#is satisfied.

Then, the centre is #(-g,-f)=(0,-4)# and radius #=sqrt(g^2+f^2-c)=sqrt10.#

#II^nd# METHOD

Completing the squares of the given eqn., we have,
#x^2+y^2+8y+16-10-=0.#
#:. (x-0)^2+(y+4)^2=10=(sqrt10)^2.#......(1)

Now recall that eqn. of a circle with centre#(h,k)# & radius #=r#, is given by,

#(x-h)^2+(y-k)^2=r^2.#............(2)

Comparing (1) & (2), we get the cetre at (0,-4)# and #r = #sqrt10.#