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What is the centre of a circle and its radius on the Cartesian plane when its equation is x squared plus y squared equals 12x -10y -12 showing work?
Wiki User
∙ 7y ago
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
Wiki User
∙ 7y ago
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What is the centre and radius of a circle embeded on the Cartesian plane whose equation is x squared plus y squared -4x -2y -4 equals 0 showing key aspects of work?
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
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
Where is the center of the circle given by the equation x plus 5 squared plus y minus 3 squared equals 25?
The centre is (-5, 3)
What is the centre and radius of a circle whose equation is x squared plus y squared -4x -6y -3 equals 0 showing work?
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
How do you solve x squared plus y squared equals 13?
The equation describes a circle with its centre at the origin and radius = √13. Each and every point on that circle is a solution.
What is the centre and radius of a circle embeded on the Cartesian plane whose equation is x squared plus y squared -4x -2y -4 equals 0 showing key aspects of work?
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
What is the Cartesian equation of a circle whose centre is in the 1st quadrant with a radius of 7 and touches the x and y axes?
Centre of the circle is at (7, 7) and its Cartesian equation is (x-7)^2 + (y-7)^2 = 49
What is the equation for a circle?
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
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
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
What is the equation and area of a circle whose diameter end points are at 2 -3 and 8 7 on the Cartesian plane showing work?
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
Where is the center of the circle given by the equation x plus 5 squared plus y minus 3 squared equals 25?
The centre is (-5, 3)
What is the centre and radius of the circle whose equation is x squared plus y squared -4x -6y -3 equals 0 showing workings?
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
What is the centre and radius of a circle whose equation is x squared plus y squared -4x -6y -3 equals 0 showing work?
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
What is the equation of a circle with center (3 8) and a radius of 2?
Centre of the circle: (3, 8) Radius of the circle: 2 Cartesian equation of the circle: (x-3)^2 + (y-8)^2 = 4
How do you solve x squared plus y squared equals 13?
The equation describes a circle with its centre at the origin and radius = √13. Each and every point on that circle is a solution.
What is the equation of a circle whose centre is at 3 -5 and meets the point of 6 -7 on the Cartesian plane?
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
What is the equation of a circle whose endpoints of its diameter are at 2 2 and 10 -4 on the Cartesian plane showing work?
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
