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If a circle with its centre at the origin has a chord with slope m the equation of the right bisector of the chord is y equals mx is this statement true or false?
Natalie3423
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 .
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What shape can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
Which of these can not form a perpendicular bisector?
Perpendicular bisector lines intersect at right angles
What is an equation of the circle whose diameter has endpoints -3 1 and 5 -2?
The equation works out as: (x-1)2+(y+0.5)2 = 18.25 Equation of a circle: (x-a)2+(y-b)2 = radius2 whereas a and b are the coordinates of the circle's centre
What is the equation of a circle centered at the origin with a radius of 5cm?
x2 + y2 = 25 A circle with centre (xo, yo) and radius r has equation: (x - xo)2 + (y - yo)2 = r2 So with centre the origin (0, 0) and radius 5 cm, the circle has equation: (x - 0)2 + (y - 0)2 = 52 ⇒ x2 + y2 = 25
Which of these tools or constructions is used to inscribe a square inside a circle?
Perpendicular bisector.
Does the perpendicular bisector of a cord of a circle passe through the center of the circle?
Yes, the perpendicular bisector of a cord is the shortest distance from the centre of a circle to the cord.
Does the perpendiculare bisector of a cord pass through the center of the circle?
Yes. The perpendicular bisector of a chord forms a radius when extended to the centre of the circle and a diameter when extended beyond the centre to the opposite point on the circumference.
What equation represents the graph of a circle?
Equation of a circle when its centre is at (0, 0): x^2 + y^2 = radius^2 Equation of a circle when its centre is at (a, b): (x-a)^2 + (y-b)^2 = radius^2
Is a circle have perpendicular bisector?
A circle can have perpendicular bisector lines by means of its diameter.
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x2 plus 4x -18y plus 59 equals 0?
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
How do you find radius of a circle?
Draw a line from any part on the outside of a circle to the exact center of the circle. * * * * * That is fine if you know where the center is but not much use if you are just given a circle and do not know where the exact centre is. In this case: Draw a chord - a straight line joining any two points on the circumference of the circle. Then draw the perpendicular bisector of the chord. Draw another chord and its perpendicular bisector. The two perpendicular bisectors will meet at the centre.
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x2 plus 4x plus y2 -18y plus 59 equals 0?
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
What is the radius of the center circle?
Equation of a circle: (x-h)^2+(y-h)^2=r^2 k is the x-coordinate for the centre, h is the y-coordinate for the centre r=raduis if the equation is x^2+y^2=r^2, the centre of the circle is at (0,0)
How can you tell what a circles equation is?
The equation for any circle is r2 = (x-a)2 + (y-b)2 Where r is the radius and the centre of the circle is (a,b)
What is the equation of a circle with a radius of 2?
In the coordinate plane, if the centre of the circle is at (a, b) then the equation is(x - a)^2 + (y - b)^2 = 2^2
What is the circle's equation with a length of it's radius is 10?
If the centre of the circle is at the point (a, b), the equation is: (x - a)2+ (y - b)2= 100.
What is the locus equidistant from ab which are 6 inches apart and also five inches away from a?
Points equidistant from AB lie on its perpendicular bisector. Points 5 inches from A lie on the circle with centre A and radius = 5 inches. You will have two points where the perp bisector and circle intersect.
