Chapter 10 Conic Sections and Systems of Nonlinear Equations

§ 10.1 Distance and Midpoint Formulas; Circles

Distance Formula & Midpoint Formula We can use the Pythagorean theorem to develop a formula for finding the distance between two points in the rectangular coordinate system. After finding a formula for distance between two points, we can use the formula to derive a formula for the midpoint of a line segment between two points. These two formulas are in turn used to derive the formula for a circle. The rectangular coordinate system gives us a unique way of knowing a circle. It enables us to translate a circle’s geometric definition into an algebraic equation. Circles occur everywhere in nature – in ripples on water, patterns on a butterfly’s wings, on cross sections of trees. Some people consider the circle to be the most pleasing of all shapes. Let’s begin by looking at the formula for the distance between points. Look at page 732 in your text to see how the formula is derived. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 10.1

Distance Formula & Midpoint Formula The Distance Formula The distance, d, between the points and in the rectangular coordinate system is The Midpoint Formula Consider a line segment whose endpoints are and . The coordinates of the segment’s midpoint are To find the midpoint, take the average of the two x-coordinates and the average of the two y-coordinates. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 10.1

Distance Formula & Midpoint Formula EXAMPLE Given the points (-4,-1) and (2,-3) find (a) the distance between the points and (b) the midpoint of the line segment with the given endpoints. SOLUTION (a) Letting and , we obtain Use the distance formula. Substitute the given values. Perform subtractions within the grouping symbols. Square 6 and -2. Add and simplify. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 10.1

Distance Formula & Midpoint Formula CONTINUED The distance between the given points is approximately 6.32 units. (b) To find the coordinates of the midpoint, we average the coordinates of the endpoints. The midpoint of the line segment between the endpoints is (-1,-2). Blitzer, Intermediate Algebra, 5e – Slide #6 Section 10.1

The Standard Form of the Equation of a Circle Definition of a Circle A circle is the set of all points in a plane that are equidistant from a fixed point, called the center. The fixed distance from the circle’s center to any point on the circle is called the radius. The Standard Form of the Equation of a Circle The standard form of the equation of a circle with center (h, k) and radius r is Blitzer, Intermediate Algebra, 5e – Slide #7 Section 10.1

Equation of a Circle EXAMPLE Write the standard form of the equation of the circle with center (-3, 5) and radius 3. SOLUTION The center is (-3, 5). Because the center is represented as (h,k) in the standard form of the equation, h = -3 and k = 5. The radius is 3, so we will let r = 3. This is the standard form of a circle’s equation. Substitute -3 for h, 5 for k and 3 for r. Simplify. The standard form of the equation of the circle is Blitzer, Intermediate Algebra, 5e – Slide #8 Section 10.1

Equation of a Circle EXAMPLE Find the center and radius of the circle whose equation is and graph the equation. SOLUTION To graph the circle, we have to know its center, (h, k), and its radius, r. We can find the values for h, k, and r by comparing the given equation to the standard form of the equation of a circle, This is with r = 2. This is with h = -3. This is with k = 2. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 10.1

Equation of a Circle CONTINUED We see that h = -3, k = 2, and r = 2. Thus, the circle has center (h, k) = (-3, 2) and a radius of 2 units. To graph this circle, first plot the center (-3, 2). Because the radius is 2, you can locate at least four points on the circle by going out two units to the right, to the left, up, and down from the center. The points two units to the right and to the left of (-3, 2) are (-1, 2) and (-5, 2), respectively. The points two units up and down from (-3, 2) are (-3, 4) and (-3, 0), respectively. Using these points, we obtain the graph that follows. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 10.1

Equation of a Circle CONTINUED This circle is said to be “tangent” to the x-axis at (-3,0), for the circle “kisses” the x-axis there. (-3,4) (-1,2) (-5,2) (-3,2) (-3,0) Blitzer, Intermediate Algebra, 5e – Slide #11 Section 10.1

The General Form of the Equation of a Circle The general form of the equation of a circle is For the equation of a circle, there is a general form and a standard form. In the standard form, it’s easy to see the center of the circle and it’s radius. In this general form, we can only see that we have either a circle or some Degenerate case of the circle. We would need to change the form to see more. In the next example, we will do just that. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 10.1

Equation of a Circle Write in standard form and graph: EXAMPLE Write in standard form and graph: SOLUTION Because we plan to complete the square on both x and y, let’s rearrange the terms so that x-terms are arranged in descending order, y-terms are arranged in descending order, and the constant term appears on the right. This is the given equation. Rewrite in anticipation of completing the square. Complete the square on x and on y. Factor on the left and add on the right. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 10.1

Equation of a Circle CONTINUED This last equation is in standard form. We can identify the circle’s center and radius by comparing this equation to the standard form of the equation of a circle, This is with r = 7. This is with h = 2. This is with k = 6. We use the center, (h, k) = (2, 6), and the radius, r = 7, to graph the circle. The graph is shown below. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 10.1

Equation of a Circle CONTINUED 7 7 7 (2,6) 7 Blitzer, Intermediate Algebra, 5e – Slide #15 Section 10.1

In conclusion… Some questions about circles… A circle is the set of all points that are equidistant from a fixed point called the center of the circle. Question… Is the center a part of the circle? Answer – no. The circle itself is just the set of points that are equidistant from the center. The distance from the center to any point on the circle is the radius of the circle. Question… Could the radius be 0? Answer – yes. In that case, what points would make up the circle? In the case that the radius is 0, and you moved out zero from the center point to the circle, the circle would be just the center point. This is the only case when the center would be a part of the circle. Question… Could the radius be a negative number? Answer – no. Question… How do you convert from the general form of the equation of a circle to the standard form? Answer… by completing the square. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 10.1

Distance Formula & Midpoint Formula Now…. Maybe you should just take a break, go outside, and find some circles in nature. It’s a beautiful world… Blitzer, Intermediate Algebra, 5e – Slide #17 Section 10.1