Find the centre and the radius of the circle | vedclass

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(A) The given equation is $x^{2}+y^{2}+8x+10y-8=0$.Rearranging the terms,we get $(x^{2}+8x) + (y^{2}+10y) = 8$.Completing the squares,we add $(8/2)^{2

Vedclass · Accepted Answer

Centre: $(-4, -5)$,Radius: $7$

Vedclass · Answer

(A) The given equation is $x^{2}+y^{2}+8x+10y-8=0$.Rearranging the terms,we get $(x^{2}+8x) + (y^{2}+10y) = 8$.Completing the squares,we add $(8/2)^{2} = 16$ and $(10/2)^{2} = 25$ to both sides:$(x^{2}+8x+16) + (y^{2}+10y+25) = 8+16+25$.This simplifies to $(x+4)^{2} + (y+5)^{2} = 49$.Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,where $(h, k)$ is the centre and $r$ is the radius:$h = -4$,$k = -5$,and $r^{2} = 49$,so $r = 7$.Thus,the centre is $(-4, -5)$ and the radius is $7$.

Find the centre and the radius of the circle $x^{2}+y^{2}+8x+10y-8=0$.

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