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In these lessons, we'll learn how to find and analyze the equations which represent these curves.
We begin with the circle.
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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A circle is a path or locus of points equidistant from a center point.
The distance between each point on the circumference and the center is called the radius.
Using this definition, we find the equation of a circle once we know the coordinates of the center and the measure of the radius.
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Let's begin with the equation of a circle whose center is at the origin.
We will do as we did when finding the equation of a line. We represent any point P by (x, y) and write an algebraic statement that says that:
the square of the distance from (x, y) to (0, 0) equals the square of the radius measure.
The reason we make the statement about the square of the radius rather than the radius, is to eliminate that big, ugly square root sign which would result if we said that the distance between (x, y) and (0, 0) was equal to the radius. Remember that the distance formula involves a square root, so to eliminate it, we write the equation of the circle as a statement about the square of the radius.
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| The equation of the circle center (0, 0), radius r is x2 + y2 = r2 |
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Example
Write the equation of the circle with center at the origin, radius = 2.
Solution: x2 + y2 = 22 u x2 + y2 = 4
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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Now let's find the equation of the a circle whose center is at a point other than the origin. So, let's find the equation of the circle with center (2, 3) and radius 4 and then we will generalize the equation we find to apply to a circle with center (h, k) and radius r.
We will write an equation that states that the square of the distance between (x, y) and (2, 3) equals the square of the radius (4).
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The equation of the circle is:
(x - 2) 2 + (y - 3) 2 = 4 2 or
(x - 2) 2 + (y - 3) 2 = 16
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Now we apply the same technique to find the equation of the circle center (h, k), radius = r :
(x - h)2 + (y - k)2 = r2
This is the standard form of the equation of the circle center (h, k), radius = r.
| The standard form of the equation of the circle with center (h, k) and radius = r is (x - h)2 + (y - k)2 = r2 |
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Example
Write the equation of the circle in standard form:
| a) center (-1, 3), radius = 5 | b) center (3, -5), radius = 1 | c) center ( -2, -6), radius = 2. |
| (x + 1)2 + (y - 3)2 = 25 | (x - 3)2 + (y + 5)2 = 1 | (x + 2)2 + (y + 6)2 = 4 |
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Example
Write the equation of the circle in standard form:
(a) with diameter endpoints at A(5, -3) and B(-3, 7).
(b) with center (-2, 1), tangent to the y-axis.
Solution:
(a) the center is the midpoint of AB which is C(1, 2)
the radius is the distance from A to C which is
the equation of the circle center (1, 2) with radius =
is:
(x - 1)2 + (y - 2)2 = 41
(b) since the circle is tangent to the y-axis, its radius = the distance between (-2, 1) and the y-axis, so the radius = 2 thus the equation is (x + 2)2 + (y - 1)2 = 4.
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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The General Form of the Equation of a Circle
The general form of the equation of a circle is derived from the standard form by squaring the binomials and collecting up the constants. Let's find the general form of the equation of the circle in part (b) of the previous Example. Here's how we do it:
(x + 2)2 + (y - 1)2 = 4 u x2 + 4x + 4 + y2 - 2y + 1 = 4
x2 + y2 + 4x - 2y + 1 = 0
As you can see, the constant term is found by taking h2 + k2 - r2 and the whole equation is set equal to zero just as we did with the general form of the equation of a line.
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| The general form of the equation of the circle is x2 + y2 - 2hx - 2ky + c = 0 where (h, k) is the center and c = h2 + k2 - r2. |
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It should be obvious then, that, when given the equation of a circle in general form, and if the coefficients of x 2 and y 2 = 1, we can find the coordinates of the center point by taking negative one half of the coefficients of x and y since they are -2h and -2k respectively. We can then find the radius by plugging the values of h and k into the formula c = h2 + k2 - r2 and solving for r.
If the coefficients of x 2 and y 2 are not = 1, they will be the same -- since that's required for the equation to represent a circle. We just divide through by this value -- and then we can find h and k by the method above.
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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Example
Find the center and radius of the circle with equation x2 + y2 + 8x - 2y + 15 = 0.
Solution:
-2h = 8, so h = -4, so, -2k = -2, which makes k = 1 ‹ the center is (-4, 1).
now c = h2 + k2 - r2 so 15 = 16 + 1 - r2 u r2 = 2 u r =
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so this circle's equation in standard form is (x + 4)2 + (y - 1)2 = 2
Another way to change the equation from general form to standard form is to complete the squares on both the x and y variables like this:
x2 + y 2 + 8x - 2y + 15 = 0 u (x 2 + 8x + 16) - 16 + (y 2 - 2y + 1) - 1 + 15 = 0
(x + 4) 2 + (y - 1) 2 - 2 = 0 u (x + 4) 2 + (y - 1) 2 = 2
As we see, the center coordinates are (-4, 1), the radius is
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Example
Put x 2 + y 2 - 14x + 4y - 11 = 0 in standard form and find the center and radius.
Solution:
Completing the squares, we get: (x2 - 14x + 49) - 49 + (y2 + 4y + 4) - 4 - 11 = 0
which becomes (x - 7)2 + (y + 2)2 = 64 so the center is (7, -2), and the radius = 8.
Example
Write the equation of the circle with center (-1, 4) passing through (3, -1) both in standard and general form.
Solution:
Since the circle passes through (3,-1), its radius = the distance between (3,-1) and (-1,4).
So, r =
‹ the equation of the circle is (x + 1)2 + (y - 4)2 = 41 in standard form
and x2 + y2 + 2x - 8y - 24 = 0 in general form.
Example
Put 3x2 + 3y2 - 12x + 18y - 69 = 0 in standard form. Find the center and radius.
Solution:
Since the coefficient of both the squared terms = 3, this is a circle, but we must divide through by 3 to use the method above.
x2 + y2 - 4x + 6y - 23 = 0
center at (2, -3) so h = 2, k = -3 and c = -23
c = h2 + k2 - r2 , substituting our values, we get -23 = 4 + 9 - r 2 u r 2 = 36
The center is at (2, -3), the radius = 6
(x - 2)2 + (y + 3)2 = 36
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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1) Write the equations of these circles in standard form:
| a) center (0, 0), radius 4 | b) center (0, 0), radius 7 | c) center (3, 0), radius 5 |
| d) center (0, -1), radius 3 | e) center (-2, 4), radius
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f) center (-5,-1), radius
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2) Write the equations of the four circles with a radius of 4 which are tangent to both axes.
3) Find the radius and the equation of the circle:
(a) center (-3, 4) passing through the origin
(b) center (4, 9) passing through point (-2, 1)
(c) with diameter endpoints at (5, -7) and (1, -3)
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4) Find the center and radius of the following circles:
| a) x2 + y2 = 36 | b) (x - 6)2 + (y + 1)2 = 16 | c) x2 + y2 - 2x + 8y + 13 = 0 |
| d) x2 + y2 + 14x + 24 = 0 | e) x2 + y2 - 6y = 0 | f) x2 + y2 - 4x + 10y + 29 = 0 |
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4) (g) write the equations of the circles in parts (c) through (e) in standard form by completing the squares.
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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1) Write the equations of these circles in standard form:
| a) center (0, 0), radius 4 x2 + y2 = 16 |
b) center (0, 0), radius 7 x2 + y2 = 49 |
c) center (3, 0), radius 5 (x - 3)2 + y2 = 25 |
| d) center (0, -1), radius 3 x2 + (y + 1)2 = 9 |
e) center (-2, 4), radius
 (x + 2)2 + (y - 4)2 = 10 |
f) center (-5,-1), radius
 (x + 5)2 + (y + 1)2 = 24 |
2) Write the equations of the four circles with a radius of 4 tangent to both axes.
In Quadrant I, the center will be (4, 4) so the equation is (x - 4)2 + (y - 4)2 = 16.
In Quadrant II, the center will be (-4, 4) so the equation is (x + 4)2 + (y - 4)2 = 16.
In Quadrant III, the center will be (-4, -4) so the equation is (x + 4)2 + (y + 4)2 = 16.
In Quadrant IV, the center will be (4, -4) so the equation is (x - 4)2 + (y + 4)2 = 16.
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3) Give the radius and the equation of the circle:
(a) center (-3, 4) passing through the origin
The radius = the distance from the origin to (-3, 4) = 5
The equation of the circle therefore is: x2 + y2 = 25
(b) center (4, 9) passing through point (-2, 1)
The radius = the distance from (4, 9) to (-2, 1) = 10
The equation of the circle therefore is: (x - 4)2 + (y - 9)2 = 100
(c) with diameter endpoints at (5, -7) and (1, -3)
The center is the midpoint of the line segment from (5, -7) to (1, -3) = (3, -5)
The radius = the distance from (3, -5) to (5, -7) =
The equation of the circle therefore is: (x - 3)2 + (y + 5)2 = 8.
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4) Find the center and radius of the following circles:
| CIRCLE | CENTER | RADIUS |
| a) x2 + y2 = 36 | (0, 0) | r = 6 |
| b) (x - 6)2 + (y + 1)2 = 16 | (6, -1) | r = 4 |
| c) x2 + y2 - 2x + 8y + 13 = 0 | (1, -4) | r = 2 |
| d) x2 + y2 + 14x + 24 = 0 | (-7, 0) | r = 5 |
| e) x2 + y2 - 6y = 0 | (0, 3) | r = 3 |
| f) x2 + y2 - 4x + 10y + 29 = 0 | (2, -5) | r = 0 (not a circle). |
| g/c) (x - 1)2 + (y + 4)2 = 4 | g/d) (x + 7)2 + y2 = 25 | g/e) x2 + (y - 3)2 = 9 |
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| intro | circle center (0, 0) | circle center (h, k) | general form |
| examples | practice | solutions |
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