Mathematics
MarisolPiatz462
12

What is the general form of the equation of the given circle with center A? CenterA:(-3,12) Radius:5 x2 + y2 + 6x − 24y − 25 = 0 x2 + y2 − 6x + 24y + 128 = 0 x2 + y2 + 6x – 24y + 128 = 0 x2 + y2 + 6x − 24y + 148 = 0

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(1) Answers
SixtaWykle

Standard form of a circle is: (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius . so we have, (x-h)^2 + (y-k)^2 = r^2 put the center (-3,12) and radius 5 (x-(-3))^2 + (y-12)^2 = 5^2 (x+3)^2 + (y-12)^2 = 25 (general form) . expand to get "general form" x^2 + y^2 + Dx + Ey + F = 0 (x+3)^2 + (y-12)^2 = 25 (x+3)(x+3) + (y-12)(y-12) = 25 x^2+6x+9 + y^2-24y+144 = 25 x^2+y^2+6x-24y+153 = 25 x^2+y^2+6x-24y+128 = 0 (answer)

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