C = pi x D 59 x 3.1416 = 185.3539 mm
59
59 x 85 = 5015.
59 x 5
59/3 times 3 equals 59
Radius = 1/2 (diameter) = 59/2 = 29.5 yards = 29 yards 1 foot 6 inches
The radius of a circle with a circumference of 59 is: 9.39
Radius of circle: 371/2*pi = 59 units to the nearest whole number
59 pi = 182.212 (rounded)
Equation of circle: x^2 +4x +y^2 -18y +59 = 0 Completing the squares: (x +2)^2 +(y -9)^2 = 26 which is radius squared Center of circle: (-2, 9) Area of circle: pi times 26 = 81.681 square units to three decimal places
C = pi x D 59 x 3.1416 = 185.3539 mm
x² + y² - 10 = 49 → x² + y² = 59 = (x - 0)² + (y - 0)² = (√59)² → circle has centre (0, 0) - the origin - and radius √59 The point (7, -2) has a distance from the centre of the circle of: √((7 - 0)² + (-2 - 0)²) = √(7² + (-2)²) = √(49 + 4) = √53 < √59 Which means that the point is INSIDE the circle and all lines drawn from it to a point on the circumference will NOT be a tangent - the lines will CROSS the circumference, not touch it. Thus there is no solution to the problem as posed. -------------------------------------------------------------------- If the equation for the circle is wrong (which is most likely given as how it was stated) please re-submit your question with the correct equation for the circle.
What equals 59 from multiplying in the 3
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
60 - 1 = 59
x² + 4x - 18y + 59 = 0 is not a circle; it can be rearranged into: y = (x² + 4x + 59)/18 which is a parabola. You have missed out a y² term. ------------------------------------------------------------ Assuming you meant: x² + 4x + y² - 18y + 59 = 0, then: The perpendicular bisector of a chord passes through the centre of the circle. The slope m' of a line perpendicular to another line with slope m is given by m' = -1/m The chord y = x + 5 has slope m = 1 → the perpendicular bisector has slope m' = -1/1 = -1 A circle with centre Xc, Yc and radius r has an equation in the form: (x - Xc)² + (y - Yc)² = r² The equation given for the circle can be rearrange into this form by completing the square in x and y: x² + 4x + y² - 18y + 59 = 0 → (x + (4/2))² - (4/2)² + (y - (18/2))² - (18/2)² + 59 = 0 → (x + 2)² +(y - 9)² - 2² - 9² + 59 = 0 → (x + 2)² + (y - 9)² = 4 + 81 - 59 → the circle has centre (-2, 9) (The radius, if wanted, is given by r² = 4 + 81 - 59 = 36 = 6²) The equation of a line with slope m' through a point (Xc, Yc) has equation: y - Yc = m'(x - Xc) → y - 9 = -1(x - -2) → y - 9 = -x - 2 → y + x = 7 The perpendicular bisector of the chord y = x + 5 within the circle x² + 4x + y² - 18y + 59 = 0 is y + x = 7
68 + 59 = 127