r= 4 (h,k) = (2,-3)?

Question

write the standard form of the equation and the general form of the equation of the circle radius r and center (h,k)

Answer 1

(x-h)^2 + (x-k)^2 = r^2
Th

Answer 2

(x - h)^2 + (y - k)^2 = r^2

(x - 2)^2 + (y -(-3))^2 = 16
(x - 2)^2 + (y + 3))^2 = 16

You can also expand the equation:

x^2 - 4x + 4 + y^2 + 6y + 9 = 16
x^2 - 4x + 4 + y^2 + 6y + 9 = 16
x^2 + y^2 - 4x + 6y + 13 = 16

x^2 + y^2 - 4x + 6y - 3 = 0

Often times you'll start out with the above, and have to put it in standard form, which involves grouping terms and completing the square for each variable

Answer 3

Standard Form: (x-2)^2 + (y+3)^2 = 4^2

General Form: x^2 + y^2 + (-4)x + 6y + (-3) = 0

Answer 4

(x - h)^2 + (y - k)^2 = r^2

(x - 2)^2 + (y - (-3))^2 = 4^2

(x - 2)^2 + (y + 3)^2 = 16

Answer 5

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Answer 6

impossible!!