Given: x2 + y2 - 10x + 4y + 4 = 0 First, we'll move our constants to the right: x2 + y2 - 10x + 4y = -4 Then group terms with the same variables together: x2 - 10x + y2 + 4y = -4 Then complete the squares: x2 - 10x + 25 + y2 + 4y + 4 = -4 + 25 + 4 (x - 5)2 + (y + 2)2 = 25 And there we have it. This is an equation for a circle whose center point is at (5, -2), with a radius of √25, which equals 5.
When y=4, x=8/5
hey,the graph of x=5 is a straight line parallel to Y-axis. all the points on the line have x-coordinate equal to 5. so on the X-axis just draw a perpendicular line 5 units away.
-2x=2y+5 +2x -2y -2y=2x+5 /-2 y=-1/1+2.5
5, of course!
5. A circle with centre (0,0) has equation: x2 + y2 = radius2 With: x2 + y2 = 25 = 52 The radius is 5.
Select any tow values of x, say x1 and x2.Calculate y1 = 5*x1 - 12 and y2 = 5*x2 - 12. Mark the points (x1, y1) and (x2, y2). Join them with a straight line and extend in both directions.
No.
The greatest possible slope is 1.
y= -5
80
y2 - 15 = 10Add 15 to each side:y2 = 25Take the square root of each side:y = +5y = -5
Centre = (0,0), radius = sqrt(5)
x2 + 10x + y2 + 20y - 19 = 0 x2 + 10x + y2 + 20y = 19 Complete the squares on the LHS: x2 + 10x + 25 + y2 + 20y + 100 = 19 + 25 + 100 = 144 Simplify LHS: (x + 5)2 + (y + 10)2 = 122 This is the equation of all points whose distance form the point (-5, -10) is 12 units. That is, the circle with centre (-5, -10) and radius = 12 units.
If x equals a constant number, the graph will be a vertical line. For example, the graph of x = 5 would be a vertical line that goes through the point (5,0). x equals 5 on every point along this lines.
x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10