recall that x^2 + y^2 = r^2 is a circle with center at (0,0) and radius r.
Now you have, since the radius is 5,
x^2 + y^2 - 6x + 4y = 25
x^2-6x+9 + y^2+4y+4 = 25+9+4
(x-3)^2 + (y+2)^2 = 38
so, k = -38
find the equation of a circle with center (0,0) radius 20.b)if the radius of a circle is x®2+y®2-6x+4y+k=0 is 5units find the constant k
2 answers
(x- xc)^2 + (y-yc)^2 = r^2
xc = yc =0 for center
x^2 + y^2 = 400
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x^2 - 2 x xc + xc^2 + y^2 - 2 y yc y^2 = r^2
here
-2 x xc = -6 x so xc = 3
-2 y yc = 4 y so yc = -2
form then is
(x - 3)^2 + (y + 2)^2 = r^2 = 25
x^2 - 6 x + 9 + y^2 + 4 y + 4 = 25
x^2 - 6 x + y^2 + 4 y + 13 = 25
x^2 - 6 x + y^2 + 4 y - 12 = 0
I guess k is -12
xc = yc =0 for center
x^2 + y^2 = 400
=======================
x^2 - 2 x xc + xc^2 + y^2 - 2 y yc y^2 = r^2
here
-2 x xc = -6 x so xc = 3
-2 y yc = 4 y so yc = -2
form then is
(x - 3)^2 + (y + 2)^2 = r^2 = 25
x^2 - 6 x + 9 + y^2 + 4 y + 4 = 25
x^2 - 6 x + y^2 + 4 y + 13 = 25
x^2 - 6 x + y^2 + 4 y - 12 = 0
I guess k is -12