0
What is the perpendicular bisector equation of the line y equals x plus 5 spanning the circle x2 plus 4x plus y2 -18y plus 59 equals 0?
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
What else can I help you with?
Is a circle have perpendicular bisector?
A circle can have perpendicular bisector lines by means of its diameter.
What can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
Is a circle can not form a perpendicular bisector?
A circle itself does not form a perpendicular bisector because a perpendicular bisector is a line that divides a segment into two equal parts at a right angle, typically associated with straight segments. However, the concept of a perpendicular bisector can be applied to chords within a circle. The perpendicular bisector of a chord will always pass through the center of the circle.
In a circle the perpendicular bisector of a chord must pass through the center of the circle?
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
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
Is a circle have perpendicular bisector?
A circle can have perpendicular bisector lines by means of its diameter.
What can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
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
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.
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
Which of these tools or constructions is used to inscribe a square inside a circle?
Perpendicular bisector.
How do you find center of a circle given 3 points on the circle?
You have points A, B, and C. Using a compass and straight edge, find a perpendicular bisector of AB (that is, a line that is perpendicular to AB and intersects AB at the midpoint of AB. Next, find a perpendicular bisector of BC. The two lines you found will meet at the center of the circle.
What is a method for constructing a regular quadrilateral?
Start with constructing a circle, then make a diameter from that circle. After you've done that, construct the perpendicular bisector of, the diameter, then draw the line in from the perpendicular bisector. After you've done that, connect the 4 points you have on the circle... then you're done. ^^ Hope this helps. :)
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x squared plus 4x plus y squared -18y plus 59 equals 0 showing work?
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
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x2 plus 4x plus y2 minus 18y plus 59 equals 0?
Chord equation: y = x+5 Circle equation: x^2 +4x +y^2 -18y +59 = 0 Both equations intersect at: (-1, 4) and (3, 8) which are the endpoints of the chord Midpoint of the chord: (1, 6) Slope of chord: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -(x-1) => y = -x+7
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
