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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?
Wiki User
∙ 7y ago
Centre of the circle is at (7, 7) and its Cartesian equation is (x-7)^2 + (y-7)^2 = 49
Wiki User
∙ 7y ago
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When was Quadrant Shopping Centre created?
Quadrant Shopping Centre was created in 1979.
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
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
Which of the following circles lie completely on the fourth quadrant?
The one in which the centre is in the fourth quadrant, and where the distance from the centre of the circle to the origin is greater than its radius.
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
To find a center in a quadrant?
It is not possible. A quadrant extends infinitely far in two directions and so has no centre.
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
What is the radius of a circle and its equation when its diameter touches the points of 2 -3 and 8 7 on the Cartesian plane showing key stages?
Points: (2, -3) and (8, 7) Centre: (8+2)/2 and (7-3)/2 = (5, 2) Radius: (8-5)2+(7-2)2 = 34 and the square root of this is the radius Equation: (x-5)2+(y-2)2 = 34
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 equation of the straight line that extends from the origin to the centre of the points 3 2 and 5 -1 on the Cartesian plane giving thorougher details of your work?
The equation is y = 1/8x because there is no y intercept and by doing some homework you'll find it correct
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
