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What is the centre and radius of a circle whose equation is x squared plus 4x plus y squared plus 6y -17 equals 0?
A circle with centre (X, Y) and radius r has an equation of the form:
(x - X)² + (y - Y)² = r²
Completing the square will allow the given equation to be rearranged into this form:
x² + 4x + y² +6y - 17 = 0
→ x² + 4x + (4/2)² - (4/2)² + y² + 6y + (6/2)² - (6/2)² - 17 = 0
→ (x + 4/2)² - 2² + (y + 6/2)² - 3² - 17 = 0
→ (x + 2)² - 4 + (y + 3)² - 9 - 17 = 0
→ (x - -2)² + (y - -3)² = 17 + 9 + 4
→ (x - -2)² + (y - -3)² = 30 = (√30)²
→ the circle has centre (-2, -3) and radius √30 (≈ 5.477).
What else can I help you with?
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.
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 when x squared plus y squared -2ax -2ay equals 0?
The centre is (a, a) and the radius is a*sqrt(2).
What is the radius for a circle with the equation x2 plus y2 equals 9?
Equation of a circle centre the origin is: x2 + y2 = radius2 ⇒ radius = √9 = 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
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.
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 when x squared plus y squared -2ax -2ay equals 0?
The centre is (a, a) and the radius is a*sqrt(2).
What is the radius for a circle with the equation x2 plus y2 equals 9?
Equation of a circle centre the origin is: x2 + y2 = radius2 ⇒ radius = √9 = 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 4 squared plus y plus 6 squared equals 49?
(-4,-6)
What is the centre and radius of a circle when x squared plus y squared -4x -2y -4 equals 0?
Equation: x^2 +y^2 -4x -2y -4 = 0 Completing the squares: (x-2)^2 +(y-1)^2 -4-1-4 = 0 So: (x-2)^2 +(y-1)^2 = 9 Therefore the centre of the circle is at (2, 1) and its radius is 3
Does the equation a squared plus b squared equals c squared classify as a quadratic equation?
no
Where is the center of the circle given by the equation x - 3 squared plus y plus 2 squared equals 9?
The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).
How do you graph the equation x squared plus y squared equals 16?
You are describing a circle, with its center at the origin and a radius of 4 (the square root of 16)
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
Is c equals 2πr the same as c equals πr squared?
No. C = 2*pi*r is the equation representing the circumference of a circle. The area of a circle is equal to pi*(r^2).
