Form a simultaneous equation with chord and circle and by solving it:- Chord makes contact with circle at: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Slope of chord: 1 Slope of perpendicular bisector: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Chord equation: y = x+5 Circle equation: x^2 +4x +y^2 -18y +59 = 0 Chord end points: (-1, 4) and (3, 8) Chord midpoint: (1, 6) Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7
Points: (1, 2) and (9, 6) Midpoint: (5, 4) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-4 = -2(x-5) => y = -2x+14 Therefore: k = -2 thus satisfying the given bisector equation
Form a simultaneous equation with chord and circle and by solving it:- Chord makes contact with circle at: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Slope of chord: 1 Slope of perpendicular bisector: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Chord equation: y = x+5 Circle equation: x^2 +4x +y^2 -18y +59 = 0 Chord end points: (-1, 4) and (3, 8) Chord midpoint: (1, 6) Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7
Points: (1, 2) and (9, 6) Midpoint: (5, 4) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-4 = -2(x-5) => y = -2x+14 Therefore: k = -2 thus satisfying the given bisector equation
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
Equation of circle: x^2 +4x +y^2 -18y +59 = 0 Completing the squares: (x+2)^2 +(y-9)^2 = 26 Equation of chord: y = x+5 Endpoints of chord: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Center of circle: (-2, 9) Slope of chord: 1 Slope of radius: -1 Perpendicular bisector equation of chord: y-6 = -1(x-1) => y =-x+7
True. (Apex)
False. 1). The proposed equation y=mx suggests that the chord's right bisector has no y-intercept, i.e. passes through the origin. This is interesting, and appears plausible, and I'm willing to acknowledge that this aspect of it is true. But ... 2). If the slope of the chord is 'm', then the slope of its right bisector is not also 'm'. If it were, that would make the chord and its bisector parallel, which would be pretty silly. The slope of any line perpendicular to the chord, including its right bisector, has to be '-1/m'. The equation of the chord's right bisector is: Y = -X/m .
Their values work out as: a = -2 and b = 4
If: y = 5x +10 and y = x^2 +4 Then: x^2 +4 = 5x +10 Transposing terms: x^2 -5x -6 = 0 Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x = -1 Therefore by substitution endpoints of the line are at: (6, 40) and (-1, 5) Midpoint of line: (2.5, 22.5) Slope of line: 5 Perpendicular slope: -1/5 Perpendicular bisector equation: y-22.5 = -1/5(x-2.25) => 5y = -x+114.75 Perpendicular bisector equation in its general form: x+5y-114.75 = 0
If: y = x + 5 Then: y² = (x + 5)² = x² + 10x +25 If: x² + 4x + y² - 18y +59 = 0 Then: x² + 4x + x² +10x + 25 - 18x - 90 + 59 = 0 Collecting like terms: 2x² - 4x - 6 = 0 Dividing all terms by 2: x² - 2x - 3 = 0 Factorizing gives: (x +1)(x -3) = 0 meaning x = -1 or 3 By substitution endpoints of chord are at: (-1, 4) and (3, 8) Midpoint of chord: (1, 6) Slope of chord: 1 Perpendicular slope of chord: -1 Perpendicular bisector equation of chord: y - 6 = -1(x -1) → y = -x + 7 Perpendicular bisector equation of chord in its general form: x + y - 7 = 0 ------------------------------------------------- The perpendicular bisector of a chord passes through the centre of the circle. The slope of the perpendicular bisector of the chord m' is such that when multiplied by the slope of the chord m the result it -1, ie m'm = -1 → m' = -1/m A circle with centre (X, Y) and radius r has form: (x - X)² + (y - Y)² = r² The equation of the circle can be rearranged into this form by completing the square in each of 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 - 2² + (y - 9)² - 9² + 59 = 0 → (x + 2)² +(y - 9)² - 4 - 81 + 59 = 0 → (x - (-2))² + (y - 9)² = 26 → The circle has centre (-2, 9) A line with an equation of the form y = mx + c has slope m The chord has equation y = x + 5 → The slope of the chord m = 1 → The slope of the perpendicular bisector m' = -1/1 = -1 A line through point (X, Y) with slope m' has equation y - Y = m'(x - X) → The equation of the perpendicular bisector is: y - 9 = -1(x - (-2)) → y - 9 = -x - 2 → y + x = 7 or: y + x - 7 = 0