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
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
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
The centre is (-5, 3)
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
The equation describes a circle with its centre at the origin and radius = √13. Each and every point on that circle is a solution.
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
Centre of the circle is at (7, 7) and its Cartesian equation is (x-7)^2 + (y-7)^2 = 49
The equation for a circle of radius r and centre (h, k) is: (x - h)² + (y - k)² = r² If the centre is the origin, the centre point is (0, 0), thus h = k = 0, and this becomes: x² + y² = r²
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
Diameter end points: (2, -3) and (8, 7) Centre of circle: (5, 2) Length of diameter: 2 times square root of 34 Equation: (x-5)^2+(y-2)^2 = 34 which in effect is the radius squared Area in square units: 34*pi
The centre is (-5, 3)
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
Centre of the circle: (3, 8) Radius of the circle: 2 Cartesian equation of the circle: (x-3)^2 + (y-8)^2 = 4
The equation describes a circle with its centre at the origin and radius = √13. Each and every point on that circle is a solution.
Centre of circle: (3, -5) Distance from (3, -5) to (6, -7) is the square root of 13 which is the radius Equation of the circle: (x-3)^2 + (y+5)^2 = 13
Endpoints: (2, 2) and (10, -4) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to (2, 2) or (10, -4) = 5 which is the radius of the circle Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25