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Инфоурок / Математика / Презентации / Презентация на тему "Circles"

Презентация на тему "Circles"


  • Математика

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The standard form of the equation of a circle with its center at the origin i...
Let's look at the equation The center of the circle is at the origin and the...
If the center of the circle is NOT at the origin then the equation for the st...
If you take the equation of a circle in standard form for example: You can fi...
If we'd have started with it like this, we'd have to complete the square on b...
Now let's work some examples: Find an equation of the circle with center at (...
Find an equation of the circle with center at (0, 0) that passes through the...
Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in...
Find an equation of the circle with center at (8, 2) and passes through the p...
Identify the center and radius and sketch the graph: To get in standard form...
Identify the center and radius and sketch the graph: Remember the center valu...
We have to complete the square on both the x's and y's to get in standard for...
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№ слайда 2 The standard form of the equation of a circle with its center at the origin i
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The standard form of the equation of a circle with its center at the origin is Notice that both the x and y terms are squared. Linear equations don’t have either the x or y terms squared. Parabolas have only the x term was squared (or only the y term, but NOT both). r is the radius of the circle so if we take the square root of the right hand side, we'll know how big the radius is.

№ слайда 3 Let's look at the equation The center of the circle is at the origin and the
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Let's look at the equation The center of the circle is at the origin and the radius is 3. Let's graph this circle. This is r2 so r = 3 Center at (0, 0) Count out 3 in all directions since that is the radius

№ слайда 4 If the center of the circle is NOT at the origin then the equation for the st
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If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this: The center of the circle is at (h, k). The center of the circle is at (h, k) which is (3,1). Find the center and radius and graph this circle. The radius is 4 This is r2 so r = 4

№ слайда 5 If you take the equation of a circle in standard form for example: You can fi
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If you take the equation of a circle in standard form for example: You can find the center and radius easily. The center is at (-2, 4) and the radius is 2. Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2)) This is r2 so r = 2 (x - (-2)) But what if it was not in standard form but multiplied out (FOILED) Moving everything to one side in descending order and combining like terms we'd have:

№ слайда 6 If we'd have started with it like this, we'd have to complete the square on b
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If we'd have started with it like this, we'd have to complete the square on both the x's and y's to get in standard form. Group x terms and a place to complete the square Group y terms and a place to complete the square Move constant to the other side 4 4 16 16 Write factored and wahlah! back in standard form. Complete the square

№ слайда 7 Now let's work some examples: Find an equation of the circle with center at (
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Now let's work some examples: Find an equation of the circle with center at (0, 0) and radius 7. Let's sub in center and radius values in the standard form 0 0 7

№ слайда 8 Find an equation of the circle with center at (0, 0) that passes through the
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Find an equation of the circle with center at (0, 0) that passes through the point (-1, -4). The point (-1, -4) is on the circle so should work when we plug it in the equation: Since the center is at (0, 0) we'll have Subbing this in for r2 we have:

№ слайда 9 Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in
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Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in the values in standard form we have: -2 5 6

№ слайда 10 Find an equation of the circle with center at (8, 2) and passes through the p
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Find an equation of the circle with center at (8, 2) and passes through the point (8, 0). Subbing in the center values in standard form we have: 8 2 Since it passes through the point (8, 0) we can plug this point in for x and y to find r2.

№ слайда 11 Identify the center and radius and sketch the graph: To get in standard form
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Identify the center and radius and sketch the graph: To get in standard form we don't want coefficients on the squared terms so let's divide everything by 9. So the center is at (0, 0) and the radius is 8/3. 9 9 9 Remember to square root this to get the radius.

№ слайда 12 Identify the center and radius and sketch the graph: Remember the center valu
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Identify the center and radius and sketch the graph: Remember the center values end up being the opposite sign of what is with the x and y and the right hand side is the radius squared. So the center is at (-4, 3) and the radius is 5.

№ слайда 13 We have to complete the square on both the x's and y's to get in standard for
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We have to complete the square on both the x's and y's to get in standard form. Group x terms and a place to complete the square Group y terms and a place to complete the square Move constant to the other side 9 9 4 4 Write factored for standard form. Find the center and radius of the circle: So the center is at (-3, 2) and the radius is 4.

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Дата добавления 21.03.2016
Раздел Математика
Подраздел Презентации
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