The circle x² + y² + 8x + 6y + 25 = 0... Find the i) Centre of the circle ii) Radius of the circle iii) Eccentricity of the circle?

x² + y² + 8x + 6y + 25 = 0

x^2+8x + y^2+6y = -25

(x^2+8x+16)-16 + (y^2+6y+9) -9 = -25

(x^2+8x+16) + (y^2+6y+9) -25 = -25

(x+4)^2+(y+3)^2 = 0

i) Center is (-4,-3)

ii) radius is 0

iii) The eccentricity of a circle is always 0.

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x² + y² + 8x + 6y + 25 = 0

x² + 8x + y² + 6y = - 25

x² + 8x + 16 + y² + 6y = - 25 + 16

x² + 8x + 16 + y² + 6y + 9 = - 25 + 16 + 9

(x² + 8x + 16) + (y² + 6y + 9) = 0

The typical equation of a circle is: (x - xo)² + (y - yo)² = R² → where:

xo: abscissa of center → - 4 in your case

yo: ordinate of center → - 3 in your case

R: radius of circle → 0 in your case

As the radius is null, it's not a circle but a point only → (- 4 ; - 3)

x² + y² + 8x + 6y + 25 = 0

the real locus of the equation is a single point in the plane.

Center:

(-8/2, -6/2)

(-4, -3)

Radius:

r² = (8/2)² + (6/2)² - 25

r² = 0

r = 0

This is not a circle at all. It is a degenerate conic of a single point, (-4, -3).

All circles have eccentricity 0, so much as that applies to circles. However, this is not a circle.