The centre of a circle is (2, -3) and the cir | vedclass

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(C) Given centre $(h, k) = (2, -3)$ and circumference $= 10\pi$.Since the circumference of a circle is given by $2\pi r$,we have $2\pi r = 10\pi$,whic

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${x^2} + {y^2} - 4x + 6y - 12 = 0$

Vedclass · Answer

(C) Given centre $(h, k) = (2, -3)$ and circumference $= 10\pi$.Since the circumference of a circle is given by $2\pi r$,we have $2\pi r = 10\pi$,which implies $r = 5$.The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 5^2$.$(x - 2)^2 + (y + 3)^2 = 25$.Expanding the squares: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 25$.$x^2 + y^2 - 4x + 6y + 13 = 25$.$x^2 + y^2 - 4x + 6y - 12 = 0$.

The centre of a circle is $(2, -3)$ and the circumference is $10\pi$. Then the equation of the circle is

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