The centre and radius of the circle x^2+y^2+8 | vedclass

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(A) The given equation of the circle is $x^2+y^2+8x+10y-8=0$. Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get: $2g =

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$(-4,-5), 7$

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(A) The given equation of the circle is $x^2+y^2+8x+10y-8=0$. Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get: $2g = 8 \implies g = 4$ $2f = 10 \implies f = 5$ $c = -8$ The centre of the circle is $(-g, -f) = (-4, -5)$. The radius $r$ is given by $\sqrt{g^2+f^2-c}$. $r = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$. Thus,the centre is $(-4, -5)$ and the radius is $7$.

The centre and radius of the circle $x^2+y^2+8x+10y-8=0$ respectively are

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