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

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(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$. $2 \pi r = 10 \pi$ $\Rightarrow r = 5$ The centre of the c

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$. $2 \pi r = 10 \pi$ $\Rightarrow r = 5$ The centre of the circle is $(h, k) = (2, -3)$. The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. Substituting the values: $(x - 2)^2 + (y - (-3))^2 = 5^2$ $(x - 2)^2 + (y + 3)^2 = 25$ $x^2 - 4x + 4 + y^2 + 6y + 9 = 25$ $x^2 + y^2 - 4x + 6y + 13 - 25 = 0$ $x^2 + y^2 - 4x + 6y - 12 = 0$ Thus,option $C$ is correct.

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

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