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Equation of the circle: 4x^2 +4y^2 -20x +8y +9 = 0

Divide all terms by 4: x^2 +y^2 -5x +2y +2.25 = 0

Completing the squares: (x-2.25)^2 +(y+1)^2 = 5

Centre of circle: (2.25, -1)

Radius of circle: square root of 5

Q: What is the centre and radius of the circle 4x squared plus 4y squared -20x plus 8y plus 9 equals 0 showing work?

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The centre is (a, a) and the radius is a*sqrt(2).

The equation describes a circle with its centre at the origin and radius = âˆš13. Each and every point on that circle is a solution.

The centre is (-5, 3)

Note that: (x-a)2+(y-b)2 = radius2 whereas a and b are the coordinates of the circle's centre Equation: x2+y2-4x-2y-4 = 0 Completing the squares: (x-2)2+(y-1)2 = 9 Therefore: centre = (2, 1) and radius = 3

x2+y2-4x-6y-3 = 0 Using the appropriate formula it works out as:- Centre of circle: (2, 3) Radius of circle: 4.

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The centre is (a, a) and the radius is a*sqrt(2).

The equation describes a circle with its centre at the origin and radius = âˆš13. Each and every point on that circle is a solution.

The centre is (-5, 3)

Note that: (x-a)2+(y-b)2 = radius2 whereas a and b are the coordinates of the circle's centre Equation: x2+y2-4x-2y-4 = 0 Completing the squares: (x-2)2+(y-1)2 = 9 Therefore: centre = (2, 1) and radius = 3

x2+y2-4x-6y-3 = 0 Using the appropriate formula it works out as:- Centre of circle: (2, 3) Radius of circle: 4.

Equation: x^2 +y^2 -4x -6y -3 = 0 Completing the squares: (x-2)^2 + (y-3)^2 -4 -9 -3 = 0 So: (x-2)^2 +(y-3)^2 = 16 Therefore the centre of the circle is at (2, 3) and its radius is 4

Equation: x^2 +y^2 -4x -6y -3 = 0 Completing the squares: (x-2)^2 +(y-3)^2 -4 -9 -3 = 0 So: (x-2)^2 +(y-3) = 16 Therefore the centre of the circle is at (2, 3) and its radius is 4

If: x^2+y^2 = 12x-10y-12 Then: x^2+y^2-12x+10y = -12 Completing the squares: (x-6)^2+(y+5)^2 -36-25 = -12 So: (x-6)^2+(y+5)^2 = 49 Therefore the centre of the circle is at (6, -5) and its radius is 7

Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius

If: x^2+y^2 -4x -2y -4 = 0 Then by completing the squares of x and y: (x-2)^2+(y-1)^2 = 9 Therefore the centre of the circle is at (2, 1) and its radius is 3 units

area equals pi r squared therefor r squared equals area over pi. Now find square root of r squared and you have "R" (radius) = 2.821

Pi (3.14) times the radius of a circle squared, equals the circumference of a circle.