Find the centre and radius of the circle x^{2 | vedclass

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(A) The given equation of the circle is $x^{2}+y^{2}-8x+10y-12=0$. Rearranging the terms,we get: $(x^{2}-8x) + (y^{2}+10y) = 12$. Completing the squar

Vedclass · Accepted Answer

$(4, -5), \sqrt{53}$

Vedclass · Answer

(A) The given equation of the circle is $x^{2}+y^{2}-8x+10y-12=0$. Rearranging the terms,we get: $(x^{2}-8x) + (y^{2}+10y) = 12$. Completing the square for $x$ and $y$: $(x^{2}-8x+16) + (y^{2}+10y+25) = 12+16+25$. This simplifies to: $(x-4)^{2} + (y+5)^{2} = 53$. Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$,we identify the centre $(h, k) = (4, -5)$ and the radius $r = \sqrt{53}$.

Find the centre and radius of the circle $x^{2}+y^{2}-8x+10y-12=0$.

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