0
What is the centre and radius of a circle on the Cartesian plane whose equation is 100x squared plus 100y squared -120x plus 100y -39 equals 0?
A circle with centre (x0, y0) and radius r has the formula:
(x - x0)² + (y - y0)² = r²
→ x² - 2x0 x + x0² + y² - 2y0 y + y0² - r² = 0
Thus (by completing the square):
100x² + 100y² - 120x + 100y - 39 = 0
→ x² + y² - 1.2x + y - 0.39 = 0
→ x² - (2 × 0.6)x + 0.6² - 0.6² + y² - (2 × 0.5)y + 0.5² - 0.5² - 0.39 = 0
→ (x - 0.6)² + (y + 0.5)² - 0.36 - 0.25 - 0.39 = 0
→ (x - 0.6)² + ( y - -0.5)² = 1 = 1²
→ the circle has centre (0.6, -0.5) and radius of 1.
Another Answer:-
Equation: 100x^2+100y^2 -120x+100y -39 = 0
Divide all terms by 100: x^2+y^2 -6/5x+y -39/100
Completing the squares of x and y: (x-3/5)^2+(y+1/2)^2 -9/25 -1/4 -39/100 = 0
So: (x-3/5)^2+(y+1/2)^2 = 1
Therefore the centre of the circle is at (3/5, -1/2) and its radius is 1
Add your answer:
Earn +20 pts
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
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 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 centre and radius of a circle that passes through the points of 2 0 and 3 1 and -6 10 on the Cartesian plane?
The equation of the circle works out as: (x+2)^2 + (y-5)^2 = 41 The circle's centre is at: (-2, 5) Its radius is the square root of 41
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 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 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 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
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 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 centre and radius of a circle enscribed through the points of 6 2 and 8 -2 and 0 2 on the Cartesian plane?
It works out that the circle's centre is at (3, -2) and its radius is 5 on the Cartesian plane.
What is the centre and radius of a circle that passes through the points of 2 0 and 3 1 and -6 10 on the Cartesian plane?
The equation of the circle works out as: (x+2)^2 + (y-5)^2 = 41 The circle's centre is at: (-2, 5) Its radius is the square root of 41
What is the radius of a circle and its centre that passes through the points of 5 0 and 3 4 and -5 0 on the Cartesian plane?
Points: (5, 0) and (3, 4) and (-5, 0) Equation works out as: x^2+y^2 = 25 Radius: 5 units in length Centre of circle is at the point of origin (0, 0) on the Cartesian plane.
What is the Cartesian equation of a circle whose end points of its diameter are at 10 -4 and 2 2?
End points: (10, -4) and (2, 2) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to any of its end points = 5 which is the radius Therefore the Cartesian equation is: (x-6)^2 +(y+1)^2 = 25
