Lecture Notes

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Circles and Distance

The Distance Formula

Recall that the Pythagorean Theorem states that if a, b, c are sides of a right triangle with c the hypotenuse, then

a² + b² = c²

Let (x,y) be a point in the plane. Then if we draw the triangle with one vertex at the origin, one at (x,y) and one at (x,0) then we have a right triangle which has one leg of length x and the other of length y. The length of the hypotenuse is the distance from the origin to the point (x,y). By the Pythagorean Theorem, this distance is

$The xy-axes, the point (x,y) a distance d from the origin, showing the right triangle with sides of length x, y, and d.$

$ d = \sqrt{x^2+y^2} $

Exercise:

Find the distance from the origin to the point (5,12)

We can generalize this notion to find the distance between two points (x₁,y₁) and (x₂,y₂)

We draw the triangle and see that the length of its legs equal

x₂ - x₁ and y₂ - y₁

$Graph of the right triangle with points (x1,y1), (x2,y2), and (x2,y1). It shows side lengths x2 - x2 and y2 - y2 and hypothenuse sqrare root of (x2 - x1)squared + (y2 - y1) squared.$

Therefore the distance from (x₁,y₁) to (x₂,y₂) is

$ d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} $

Example

Find the distance from (2,-3) to (-1,4)

Solution:

We use the distance formula:

$ d = \sqrt{(-1 - 2)^2+(4 - (-3))^2} = \sqrt{9+49} = \sqrt{58}$

Circles

Recall that a circle with radius r and center (h,k) is defined by the set of point of distance r from the point (h,k). If (x,y) is on the circle then the distance from (h,k) to (x,y) is r. The distance formula tells us that $Graph of the circle with center (h,k), radius r and showing the line segment from the center to the point (x,y) on the circle.$

(x - h)² + (y - k)² = r²

This is called the standard form of the equation of the circle.

Example

The equation of the circle with radius 3 centered at (1,2) is

(x - 1)² + (y - 2)² = 9

Example: Find the center and radius of the circle

x² + y² - 4x + 6y - 12 = 0

We must get the equation into standard form by completing the squares. We have

x² - 4x + y² + 6y = 12

-4/2 = -2 and 6/2 = 3

(-2)² = 4 and 3² = 9

Adding and subtracting we have

x² - 4x + 4 - 4 + y² + 6y + 9 - 9 = 12

(x - 2)² + (y + 3)² = 12 + 4 + 9 = 25 = 5²

So that the center of the circle is (2,-3) and the radius is 5.

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