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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 length of the tangent line from the coordinate of 9 0 to a point where it touches the circle of x2 plus 8x plus y2 -9 equals 0?
Circle equation: x^2 +8x +y^2 -9 = 0 Completing the square: (x+4)^2 +y^2 = 25 Center of circle: (-4, 0) Radius of circle: 5 Distance from (-4, 0) to (9, 0) = 13 which will be the hypotenuse of a right triangle Length of tangent line using Pythagoras; theorem: 13^2 -5^2 = 144 Therefore length of tangent line is the square root of 144 = 12 units
What is the length of 3 0 0 -6?
Distance between (3, 0) and (0, -6) = sqrt[(3 - 0)2 + (0 - -6)2] = sqrt(32 + 62) = sqrt(45) = 3*sqrt(5) or 6.71 approx.
What is the length of the line x -y equals 10 that spans the curve x squared plus y squared plus 4x plus 6y -40 equals 0 showing work?
If: x^2+y^2+4x+6y-40 = 0 and x-y =10 or as x = 10+y Then: (10+y)^2+y^2+4(10+y)+6y-40 = 0 Thus: 100+20y+y^2+y^2+40+4y+6y-40 = 0 Collecting like terms: 2y^2+30y+100 = 0 or as y^2+15y+50 = 0 When factored: (y+5)(y+10) = 0 So: y = -5 or y = -10 By substitution equations intersect at: (0, -10) and (5, -5) Length of line is the square root of: (5-0)^2 plus (-5--10)^2 = square root of 50 Therefore length of line is: square root of 50 or about 7.071 to three decimal places
Is 2/5 closer to 1 or 0?
2/5 is closer to 0.
