What is the center x^2+y^2-8x-6y+21=0 ?

The center x^2+y^2-8x-6y+21=0 is (4,3)

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Find the Center and Radius x^2+y^2-8x-6y+21=0

center

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

(4, 3)

Solution steps

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

Rewrite

x^{2} + y^{2} - 8 x - 6 y + 21 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 3), r = 2

And the center is:

(4, 3)

y = - 6 x^{2} + 3 x + 2

h (x) = x^{2} - 14 x + 6

8 x^{2} - 48 x + 65

f (x) = x^{2} - 14 x + 6

f (x) = x^{2} - 6 x - 4