Find the expanded form of the circle equation with center at point ((h, k) = (-3, 4)) and radius (r = 5)

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Step:1 ：We know that the standard equation of a circle is ((x-h)^2 + (y-k)^2 = r^2), where ((h, k)) is the center and (r) is the radius of the circle.Step:2 ：Substitute (h=-3), (k=4), and (r=5) into the equation to get ((x-(-3))^2 + (y-4)^2 = 5^2) or ((x+3)^2 + (y-4)^2 = 25).Step:3 ：Expand the equation to get (x^2 + 6x + 9 + y^2 - 8y + 16 = 25).Step:4 ：Simplify the equation to get (x^2 + y^2 + 6x - 8y = 0).

Answer

Step: 1 ：We know that the standard equation of a circle is ((x-h)^2 + (y-k)^2 = r^2), where ((h, k)) is the center and (r) is the radius of the circle.Step: 2 ：Substitute (h=-3), (k=4), and (r=5) into the equation to get ((x-(-3))^2 + (y-4)^2 = 5^2) or ((x+3)^2 + (y-4)^2 = 25).Step: 3 ：Expand the equation to get (x^2 + 6x + 9 + y^2 - 8y + 16 = 25).Step: 4 ：Simplify the equation to get (x^2 + y^2 + 6x - 8y = 0).

Find the expanded form of the circle equation with center at point \((h, k) = (-3, 4)\) and radius \(r = 5\)

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