How do you find the center and radius of the circle #x^2 + y^2 - 10x - 8y - 8 = 0#?

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How do you find the center and radius of the circle #x^2 + y^2 - 10x - 8y - 8 = 0#?

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Answer 1 · 2016-06-29T18:39:18

The centre is #(5,4)#, radius #r=7.#

Explanation:

The given eqn. :#x^2+y^2-10x-8y-8=0.#
Completing squares on #L.H.S.#, we get,
# x^2-10x+25+y^2-8y+16=8+25+16=49.#
#:. (x-5)^2+(y-4)^2=7^2................(1)#

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

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

Comparing #(1)# with the std. eqn, we find that,
the centre is #(5,4)#, radius #r=7.#

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How do you find the center and radius of the circle #x^2 + y^2 - 10x - 8y - 8 = 0#?

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