Answer :

[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x-9)^2+(y+4)^2=1 \\\\\\\ [x-\stackrel{h}{9}]^2+[y-(\stackrel{k}{-4})]^2=1\qquad \qquad \qquad center~(9,-4)[/tex]
abidemiokin

Answer:

(9, -4)

Step-by-step explanation:

Given the general equation of a circle to be x²+y²+2gx+2fy+c = 0 where the coordinates of the centre is give as (-g,-f)

Given the equation of the circle

(x – 9)² + (y + 4)² = 1

First, we need to expand the equation and write it in its standard form as shown above to have;

x²-18x+81+y²+8y+16 = 1

x²+y²-18x+8y+16-1 = 0

x²+y²-18x+8y+15 = 0

Comparing the resulting equation to the general equation above to get g and f,

2gx = -18x

2g = -18

g = -9

Similarly,

2fy = 8y

2f = 8

f = 4

The circle centre of the circle (-g,-f) = (9,-4)

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