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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?
Wiki User
∙ 7y ago
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
Wiki User
∙ 7y ago
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What is the equation to find the perimeter of a circle?
Circumference ("perimeter") of a circle = (pi) x (diameter of the circle)
If the radius of a circle is 8 what is the diameter?
If the radius of a circle is 8, then the diameter is 16. The diameter of a circle is two times of a radius The equation would be d = 2r so plugging in 8 for r, the equation would become d = 2(8) Therefore d is 16.
What is the equation of the circle standard form?
Area of a circle = pi*radius squared Circumference of a circle = 2*pi*radius or diameter*pi
How does the circumference of a circle change if you double its diameter?
Since the equation for circumference is pi*d then if you double the diameter the circumference will be doubled.
What is the equation for determining radius of a circle if you have the circumference?
Circumference = Diameter multiplied by Pi Diameter = Circumference divided by pi Radius = Diameter divided by 2.
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
What is the equation of a circle whose centre lies in the 1st quadrant touching both the x and y axes with a diameter of 14 inches?
If you mean as on the Cartesian plane then the following key notes are applicable:-Diameter: 14 inchesRadius: 7 inchesCentre of the circle is at: (7, 7)Equation of the circle: (x-7)^2 +(y-7)^2 = 49
If you Give a circle with a diameter of 4 what is the circumference?
the equation for the circumference of a circle is 2*radius*pi, or since the radius is half of the diameter 2*radius=diameter, we can simplify the equation to the circumference of a circle=diameter*pi
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
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 the radius of a circle?
The radius of a circle = the diameter of the circle divided by 2
What is the equation of a circle whose diameter end points are at 8 7 and 2 -3 on the Cartesian plane?
End points: (8, 7) and (2, -3) Midpoint: (5, 2) which is the center of the circle Radius: square root of 34 Equation of the circle: (x-5)^2 +(y-2)^2 = 34
What is the equation for the diamiter of a circle?
Diameter = 1/2 of the circle's radius. Diameter = 2 times the square root of (the circle's area/pi) Diameter = the circle's circumference/pi
What is the equation of a circle when the endpoints of its diameter are at 10 -4 and 2 2 on the Cartesian plane?
Endpoints: (10, -4) and (2, 2) Midpoint: (6, -1) which is the centre of the circle Distance from (6, -1) to (10, -4) or (2, 2): 5 which is the radius of the circle Therefore equation of the circle: (x-6)^2 + (y+1)^2 = 25
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
What is the equation to find the perimeter of a circle?
Circumference ("perimeter") of a circle = (pi) x (diameter of the circle)
What is the equation of a circle whose diameter endpoints are at 2 -3 and 8 7 on the Cartesian plane and what are the tangent equations touching its endpoints showing work?
Endpoints: (2, -3) and (8, 7)Centre of circle: (5, 2)Radius of circle is the square root of 34Equation of the circle: (x-5)^2 +(y-2)^2 = 34Slope of radius: 5/3Slope of tangents: -3/51st tangent equation: y--3 = -3/5(-2) => 5y = -3x-92nd tangent equation: y-7 = -3/5(x-8) => 5y = -3x+59
