FIND THE CENTRE AND RADIUS OF THE CIRCLE?

Of course its a circle, Joe. The x² and y² coefficents are equal.

x² + y² - 10x + 2y + 25 = 0

(x² - 10x + 25) + (y² + 2y + 1) - 1 = 0

(x-5)² + (y+1)² = 1

Center: (x,y) = (5,-1)

Radius: 1

X*2+Y*2-10X+2Y+25=0

you need to get in the format (x+a)^2 + (y+b)^2 = c

a and b can be ant number; c must be positive

regroup

(x^2 -10x + 25) + (y^2 + 2y) = 0

add 1 to each side to complete the square

(x^2 -10x +25) + (y^2 +2y +1) = 1

(x-5)^2 + (y+1)^2 = 1

center = 5, -1

you get that by solving (x-5) = 0

x = 5

(y+1) = 0

y = -1

radius = sqrt of the c term

radius =sqrt(c) = sqrt(1)

r = 1

x^2 + y^2 - 10x + 2y + 25 = 0

Comparing with the general equation of circle

2g = - 10

g = -5

-g = 5

2f = 2

f = 1

-f = -1

Centre of circle is at ( 5 , -1 )

Radius = √ (g^2 + f^2 - c )

= √ 25 + 1 - 25 = 1 unit

Radius = 1 unit

(x - 5)^2 - 25 + (y + 1)^2 - 1 + 25 = 0

(x - 5)^2 + (y + 1)^2 = 1

Center: (5, -1)

Radius: 1

(x² - 10x) + (y² + 2y) = - 25

(x² - 10x + 25) + (y² + 2y + 1) = - 25 + 26

(x - 5)² + (y + 1)² = 1

C (5 , - 1) , r = 1

Let us consider the general form of the equation

x2+y2+2gx+2hy+c=0.

Compare it with given equation.

Centre=(g,h).

Rad=(g2+h2-c)^1/2

2 pye d.

thats not a circle

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