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
(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
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
To find the equation of a circle with center (0,0) and radius 20, we can use the standard form of the equation for a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
In this case, the center of the circle is (0, 0) and the radius is 20. Plugging these values into the equation, we get:
(x - 0)^2 + (y - 0)^2 = 20^2
Simplifying, we get:
x^2 + y^2 = 400
Therefore, the equation of the circle with center (0, 0) and radius 20 is x^2 + y^2 = 400.
b) If the equation of the circle is x^2 + y^2 - 6x + 4y + k = 0 and the radius is 5 units, we can change the equation to the standard form as follows:
Rearrange the terms by grouping the x and y terms together:
(x^2 - 6x) + (y^2 + 4y) + k = 0
Complete the square for both the x and y terms:
(x^2 - 6x + 9) + (y^2 + 4y + 4) + k = 9 + 4
Simplify the squares:
(x - 3)^2 + (y + 2)^2 + k = 13
Comparing this to the standard form of the equation for a circle, we can see that the center of the circle is (3, -2) and the radius is the square root of the constant term on the right side of the equation, which in this case is 5 units.
Therefore, the constant k is 13 - 5 = 8.