![]() ![]() And the number must be greater than the negative of the sum of the squares of half the coefficients of x and y. ![]() We can say, then, that when a quadratic in x plus a quadratic in y is equal to a number. This is the equation of a circle of radius 4, whose center is at (2, 1). To complete the square in y, we will add 1 to both sides. To complete the square in x, we will add 4 to both sides. Therefore, we will complete the square in both x and y. To show that something is the equation of a circle, we must show that it can have this form: Name the radius and the coördinates of the center. Show that the following is the equation of a circle. Write the equation of the circle of radius 3, and center at the following point. The circle has been translated from (0, 0) to ( a, b). What is the equation of a circle with center at ( a, b) and radius r? is the equation of a circle of radius 5 centered at the origin. That is the equation of a circle of radius r, with center at the origin (0, 0). Therefore, according to the Pythagorean distance formula for the distance of a point from the origin: What, then, characterizes every point ( x, y) on the circumference of a circle?Įvery point ( x, y) is the same distance r from the center. The equation of any figure is the relationship between every coördinate pair on the figure. Transpose the constant term, and complete the square on the right: y − 1 What are the coördinates of the vertex of this parabola? The right-hand side is the perfect square of ( x − 5). ![]() y + 2Ĭomplete the square by adding 9 to both sides: we will transpose the constant term, and complete the square on the right. What are the coördinates of the vertex of this parabola? y Again, we must make the equation look like this: Now, x 2 + 6 x + 9 is the perfect square of ( x + 3): To answer, we must make the equation look like this: Write the equation of the parabola whose vertex is at a) (1, 2)Įxample 3. This is a translation of y = x 2 to ( a, b). Write the equation of the parabola (with leading coefficient 1) whose vertex is at the point ( a, b).Īnswer. This is equivalent to y − 1 = − x 2, which is the reflected parabola translated 1 unit up.Įxample 2. It has been translated 4 units left and 3 units down. To cover the answer again, click "Refresh" ("Reload"). To see the answer, pass your mouse over the colored area. The graph of the absolute value has been translated 3 units up, but 5 units to the left. In the example above, the argument of | x| becomes x − 3.Įxample 1. When f( x) is translated a units horizontally, then the argument of f( x) becomes x − a. Vertically, then the equation of the translated So the image (that is, point B) is the point (1/25, 232/25).Is translated a units horizontally and b units So the intersection of the two lines is the point C(51/50, 457/50). Now we need to find the intersection of the lines y = 7x + 2 and y = (-1/7)x + 65/7 by solving this system of equations. So the equation of this line is y = (-1/7)x + 65/7. So the desired line has an equation of the form y = (-1/7)x + b. Since the line y = 7x + 2 has slope 7, the desired line (that is, line AB) has slope -1/7 as well as passing through (2,9). So we first find the equation of the line through (2,9) that is perpendicular to the line y = 7x + 2. Then, using the fact that C is the midpoint of segment AB, we can finally determine point B.Įxample: suppose we want to reflect the point A(2,9) about the line k with equation y = 7x + 2. Then we can algebraically find point C, which is the intersection of these two lines. So we can first find the equation of the line through point A that is perpendicular to line k. Note that line AB must be perpendicular to line k, and C must be the midpoint of segment AB (from the definition of a reflection). Let A be the point to be reflected, let k be the line about which the point is reflected, let B represent the desired point (image), and let C represent the intersection of line k and line AB. ![]()
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