Find the equation of the circle whose center | vedclass

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(A) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.Given center $(h, k) = (3, 5)$ and

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$x^2 + y^2 - 6x - 10y + 18 = 0$

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(A) The standard equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.Given center $(h, k) = (3, 5)$ and radius $r = 4$.Substituting these values into the formula:$(x - 3)^2 + (y - 5)^2 = 4^2$Expanding the squares:$(x^2 - 6x + 9) + (y^2 - 10y + 25) = 16$Rearranging the terms:$x^2 + y^2 - 6x - 10y + 9 + 25 - 16 = 0$$x^2 + y^2 - 6x - 10y + 18 = 0$

Find the equation of the circle whose center is $(3, 5)$ and radius is $4$.

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