Chapter 6

Conic Sections

In this chapter, we will study the basic properties and equations of circles and parabolas.

6.1     Introduction

The curves that can be obtained by the intersections of a plane with a double right-circular cone are called conics or conic sections. The cone consists of two parts, called nappes that intersect at the vertex. The most important of conic sections or simply conics are circles, ellipses, parabolas and hyperbolas.
fig 6.1-1

fig 6.1-1b
image credit

A circle is obtained by cutting a cone with a plane that is perpendicular to the axis and does not contain the vertex. If the cutting plane is tilted slightly and intersects only one nappe, the resulting intersection is an ellipse. If the cutting plane is tilted still further so that it is parallel to a line on the surface of a cone, but intersects only one nappe, the resulting intersection is a parabola. Finally, if the plane intersects both nappes, but does not contain the vertex, the resulting intersection is a hyperbola.

All conics can be written in the form of the equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.

The equation of each of these conics can be written in a standard form which can tell the different types of conics without having to graph it.

Today, properties of conic sections are used in the construction of telescopes, radar antennas, medical equipment, navigational systems, and in the determination of satellite orbits.

6.2   Circles

A circle is the set of all points in a plane that are equidistant from a fixed point. the fixed point is called the center of the circle and the equidistant is called the radius of the circle.

Now, we consider a circle with a radius r and the center (h, k).

Standard Form of Equation of a Circle

If (h, k) is a fixed point in the plane, then the circle of radius r centered at (h, k) is the set of all points in the plane whose distance from (h, k) is r.

Thus, a point (x, y) will lie on this circle if and only if
$$\small{\sqrt{(x-h)^2 + (y-k)^2} = r \qquad (1)}$$ or equivalently,
(x - h)2 + (y - k)2 = r2           (2)
fig 6.1-2


This is called the standard form of the equation of a circle. We can find immediately the center and radius from the equation.
The center = (h, k)     and     the radius = r.

General Form of the Equation of a Circle

Expanding the equation 2, we get
x2 + y2 - 2hx - 2ky + h2 + k2 - r2 = 0
Taking 2g = - 2h, 2f = -2k, c = h2 + k2 - r2, the equation becomes the form
x2 + y2 + 2gx + 2fy + c = 0
This is called the general form of the equation of the circle.
We consider the conic equation
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
If A = C and B = 0, the equation becomes a circle equation.
Note that B = 0 means that the equation does not contain the term xy. Again, we can find the center and radius of this equation by using the technique of completing the squares.
x2 + y2 + 2gx + 2fy = -c
x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 - c
(x + g)2 + (y + f)2 = g2 + f2 - c
In this case, the center is (-g, -f) and the radius is
$$\small{\sqrt{g^2 + f^2 - c}.}$$

Degenerate Cases of a Circle

Consider the equation
x2 + y2 + 2gx + 2fy + c = 0
  • If g2 + f2 - c > 0, the graph is a circle with center at (-g, -f) and radius is $$\small{\sqrt{g^2 + f^2 - c}.}$$
  • If g2 + f2 - c = 0, the graph is a single point (-g, -f).
  • If g2 + f2 - c < 0, the equation has no real solution and consequently no graph.


    • Example 1.
    Find the equation of the circle radius 4 centered at (-1, 2) in standard form.
    Solution
    Since r = 4,
    (h - k) = (-1, 2), the equation of the circle is
    (x + 1)2 + (y - 2)2 = 42
    (x + 1)2 + (y - 2)2 = 16.

    Example 2.
    Find an equation of the circle with center (-1, 2) that passes through (3, -1).
    Solution
    Since     (h, k) = (-1, 2), (x1, y1) = (3, -1) , we have
    $$\small{ r = \sqrt{(-1 - 3)^2 + (2 + 1)^2} = 5.}$$
    Thus, the equation of the circle is
    (x + 1)2 + (y - 2)2 = 52
    x2 + y2 + 2x - 4y - 20 = 0.
    fig 6.1-3


    Example 3.
    Find the center and radius of each of the following circles.
    (a) (x - 3)2 + (y - 5)2 = 16
    (b) (x + 7)2 + (y - 2)2 = 9
    (c) x2 + y2 = 1
    (d) x2 + (y - 3)2 = 49
    Solution
    Equation of Circle Center (h, k) Radius r
    (x - 3)2 + (y - 5)2 = 16 (3, 5) 4
    (x + 7)2 + (y - 2)2 = 9 (-7, 2) 3
    x2 + y2 = 1 (0, 0) 1
    x2 + (y - 3)2 = 49 (0, 3) 7
    The circle x2 + y2 = 1 is the unit circle centered at the origin and has radius 1.

    Example 4.
    Find the center and radius of the following circles.
    (a) x2 + y2 - 2x + 10y + 23 = 0
    (b) 2x2 + 2y2 - 24x + 47 = 0
    Solution
    (a) Method 1
    x2 + y2 - 2x + 10y + 23 = 0
    x2 - 2x + y2 + 10y = -23
    x2 - 2x + 12 + y2 + 10y + 52 = 1 + 25 - 23
    (x - 1)2 + (y + 5)2 = 3
    Therefore, the center = (1, -5) and radius = √ 3

    Method 2
    x2 + y2 - 2x + 10y + 23 = 0
    Comparing with
    x2 + y2 + 2gx + 2fy + c = 0,
    we get
    2g = -2,     2f = 10,     c = 23.
    and so
    g = -1,   f = 5,   c = 23.
    Therefore, the center = (-g, -f) = (1, -5)   and
    $$\small{ \text{radius} = \sqrt{g^2 + f^2 - c}}$$ $$\small{ \sqrt{1 + 25 - 23} = \sqrt{3}.}$$
    (b) Method 1
    2x2 + 2y2 - 24x + 47 = 0
    x2 + y2 - 12x + 47 ⁄ 2 = 0
    x2 - 12x + y2 = - 47 ⁄ 2
    x2 - 12x + 62 + y2 = 36 - 47 ⁄ 2
    (x - 6)2 + y2 = 25 ⁄ 2
    (x - 6)2 + y2 = (5 ⁄ √ 2 )2
    The center = (6, 0)   and   radius = 5 ⁄ √ 2 = 5 √ 2 ⁄ 2.
    Method 2
    2x2 + 2y2 - 24x + 47 = 0
    x2 + y2 - 12x + 47 ⁄ 2 = 0
    Comparing with
    x2 + y2 + 2gx + 2fy + c = 0,
    we get
    2g = -12,   2f = 0,   c = 47 ⁄ 2
    and so
    g = -6,   f = 0
    Therefore, the center = (-g, -f) = (6, 0)   and
    $$\small{ \text{radius} = \sqrt{g^2 + f^2 - c} = \sqrt{36 + 0 - \frac{47}{2}}}$$ $$\small{ = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2} }= \frac{5 \sqrt{2}}{2}. }$$

    Example 5.
    Describe the graphs of the following equations.
    (a) (x - 1)2 + (y + 4)2 = -9
    (b) (x - 1)2 + (y + 4)2 = 0
    (c) x2 + y2 - 4x + 2y - 44 = 0
    Solution
    (a) (x - 1)2 + (y + 4)2 = -9
    We see that the right hand side is less than zero, so the equation has no graph.

    (b) (x - 1)2 + (y + 4)2 = 0
    The graph is a single point (1, -4).

    (c) x2 + y2 - 4x + 2y - 44 = 0
    Comparing with
    x2 + y2 + 2gx + 2fy + c = 0,
    we get
    2g = -4,   2f = 2,   c = -44, 
    and so
    g = -2,   f = 1,   c = -44.
    Since   g2 + f2 - c = 4 + 1 + 44 = 49 > 0,
    the graph is a circle with center at (2, -1) and radius is 7.




    Exercise 6.1

    1. Find the center and radius of each circle:
    (a) x2 + y2 = 16
    Solution
    (x - h)2 + (y - k)2 = r2
    (x - 0)2 + (y - 0)2 = 42
    center = (0, 0),     radius = 4

    (b) (x - 1)2 + (y - 4)2 = 25
    Solution
    (x - h)2 + (y - k)2 = r2
    (x - 1)2 + (y - 4)2 = 52
    center = (1, 4),     radius = 5

    (c) (x - 3)2 + (y + 5)2 = 49
    Solution
    (x - h)2 + (y - k)2 = r2
    (x - 3)2 + (y + 5)2 = 72
    center = (3, -5),     radius = 7

    (d) (x + 1)2 + y2 = 1
    Solution
    (x - h)2 + (y - k)2 = r2
    (x + 1)2 + (y - 0)2 = 12
    center = (-1, 0),     radius = 1

    (e) 2x2 + 2y2 - 12x + 8y - 24 = 0
    Solution
    2x2 + 2y2 - 12x + 8y - 24 = 0
    x2 + y2 - 6x + 4y - 12 = 0
    x2 - 6x + 9 - 9 + y2 + 4y + 4 - 4 - 12 = 0
    (x2 - 6x + 9) - 9 + (y2 + 4y + 4) - 4 - 12 = 0
    (x - 3)2 + (y + 2)2 = 25
    (x - h)2 + (y - k)2 = r2
    (x - 3)2 + (y + 2)2 = 52
    center = (3, -2),     radius = 5

    (f) x2 + y2 + x + y - 1 ⁄ 2 = 0
    Solution
    comparing with
    x2 + y2 + 2gx + 2fy + c = 0,
    we get
    2g = 1,     2f = 1,     c = - 1⁄2
    and so
    g = 1⁄2
    ,   f = 1⁄2
    ,   c = - 1⁄2
    .
    Therefore, the center = (-g, -f) = (- 1⁄2, - 1⁄2 )   and
    $$\small{\text{radius} = \sqrt{g^2 + f^2 - c}}$$ $$\small{ = \sqrt{(- \dfrac{1}{2})^2 + (- \dfrac{1}{2})^2 - (- \dfrac{1}{2})}}$$ $$\small{ = \sqrt{\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{2}}}$$ $$\small{= \sqrt{\dfrac{2}{4} + \dfrac{1}{2}}}$$ $$\small{= \sqrt{\dfrac{1}{2} + \dfrac{1}{2}} = 1}$$ 2. Find the standard equation of the circle satisfying the given conditions.
    (a) center(3, -2),   radius = 4.
    Solution
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    -h = 3, - k = -2   and r = 4
    h = -3, k = 2
    (x - 3)2 + (y + 2)2 = 42
    (x - 3)2 + (y + 2)2 = 16

    (b) center(2, 5)   radius = √ 5 .
    Solution
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    -h = 2, - k = 5   and r = √ 5
    h = -2, k = -5
    (x - 2)2 + (y - 5)2 = ( √ 5 )2
    (x - 2)2 + (y - 5)2 = 5

    (c) center (-4, 1),   radius = 3.
    Solution
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    -h = -4, - k = 1   and r = 3
    h = 4, k = -1
    (x + 4)2 + (y - 1)2 = 32
    (x + 4)2 + (y - 1)2 = 9

    (d) center(-4, 8),   circle is tangent to the x-axis.
    Solution
    Since the circle is tangent to the x-axis, its center lies on a line parallel to the x-axis, which is 8 units above the x-axis.
    (h, k) = (-4, 8),   r = 8
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    (x - (-4) )2 + (y - 8)2 = 82
    (x + 4)2 + (y - 8)2 = 64
    (-4, 8) 8 x-axis
    (e) center(5, 8),   circle is tangent to the y-axis.
    Solution
    Since the circle is tangent to the y-axis, its center lies on a line parallel to the y-axis, which is 5 units to the right of the y-axis.
    (h, k) = (5, 8),   r = 5
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    (x - 5)2 + (y - 8)2 = 52
    (x - 5)2 + (y - 8)2 = 25
    (5, 8)

    (f) center(-3, -2),   circle passes through the origin.
    Solution
    (h, k) = (-3, -2)
    $$\small{ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}} $$ $$\small{= \sqrt{(0 - (-3))^2 + (0 - (-2))^2}}$$ $$\small{= \sqrt{9 + 4} = \sqrt{13}} $$ The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    (x - (-3))2 + (y - (-2))2 = ( √ 13 )2
    (x + 3)2 + (y + 2)2 = 13
    (-3, -2) (0, 0)

    (g) center(4, -5),   circle passes through (1, 3).
    Solution
    (h, k) = (4, -5),   (x1, y1) = (1, 3)
    $$\small{ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}} $$ $$\small{= \sqrt{(1 - 4)^2 + (3 - (-5))^2}}$$ $$\small{= \sqrt{9 + 64} = \sqrt{73}} $$ The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    (x - 4)2 + (y - (-5))2 = ( √ 73 )2
    (x - 4)2 + (y + 5)2 = 73.

    (h) diameter has end points (6, 1) and (-2, 3).
    Solution
    $$\small{\text{diameter} = \sqrt{(x_2 - x_1) + (y_2 - y_1)}}$$ $$\small{\sqrt{(-2 - 6)^2 + (3 - 1)^2}}$$ = √ 64 + 4 = √ 68
    $$\small{ \text{radius} = \dfrac{\sqrt{68}}{2} = \sqrt{\dfrac{68}{4}} = \sqrt{17}}$$ The center of a circle is the midpoint of its diameter.
    M = (x1 + x2⁄2, y1 + y2⁄2 )
    = (6 - 2⁄2, 1 + 3⁄2 )
    = (2, 2)
    (h, k) = (2, 2)
    The standard form of the equation of a circle is
    (x - h)2 + (y - k)2 = r2
    (x - 2)2 + (y - 2)2 = ( √ 17 )2
    (x - 2)2 + (y - 2)2 = 17.
    (-2, 3) (6, 1)

    3. Determine whether each equation represents a circle, a point or no graph. If the equation represents a circle, find the center and radius.
    (a) x2 + y2 - 6x + 2y + 9 = 0
    Solution
    x2 + y2 - 6x + 2y + 9 = 0
    x2 - 6x + 9 - 9 + y2 + 2y + 1 - 1 + 9 = 0
    (x2 - 6x + 9) - 9 + (y2 + 2y + 1) - 1 + 9 = 0
    (x2 - 6x + 9) + (y2 + 2y + 1) - 1 = 0
    (x - 3)2 + (y + 1)2 = 1
    The equation is a circle with center at (3, -1) and radius is 1.

    (b) x2 + y2 - 4x - 6y + 13 = 0
    Solution
    x2 + y2 - 4x - 6y + 13 = 0
    x2 -4x + 4 - 4 + y2 - 6y + 9 - 9 + 13 = 0
    (x2 -4x + 4) - 4 + (y2 - 6y + 9) - 9 + 13 = 0
    (x2 -4x + 4) + (y2 - 6y + 9) - 4 - 9 + 13 = 0
    (x - 2)2 + (y - 3)2 = 0
    The equation is a point.

    (c) x2 + y2 + 10y + 26 = 0
    Solution
    x2 + y2 + 10y + 26 = 0
    x2 + y2 + 10y + 25 - 25 + 26 = 0
    x2 + (y2 + 10y + 25) - 25 + 26 = 0
    x2 + (y + 5)2 + 1 = 0
    x2 + (y + 5)2 = -1
    The right hand side is less than zero, so the equation has no graph.

    (d) 2x2 + 2y2 - 12x + 8y - 24 = 0
    Solution
    2x2 + 2y2 - 12x + 8y - 24 = 0
    x2 + y2 - 6x + 4y - 12 = 0
    x2 - 6x + 9 - 9 + y2 + 4y + 4 - 4 - 12 = 0
    (x2 - 6x + 9) - 9 + (y2 + 4y + 4) - 4 - 12 = 0
    (x - 3)2 + (y + 2)2 - 25 = 0
    (x - 3)2 + (y + 2)2 = 25
    (x - 3)2 + (y + 2)2 = 52
    The equation is a circle with center at (3, -2) and radius is 5.

    (e) x2 + y2 - 10x - 2y + 29 = 0
    Solution
    x2 + y2 - 10x - 2y + 29 = 0
    x2 - 10x + 25 -25 + y2 - 2y + 1 - 1 + 29 = 0
    (x2 - 10x + 25) -25 + (y2 - 2y + 1) - 1 + 29 = 0
    (x2 - 10x + 25) + (y2 - 2y + 1) - 25 - 1 + 29 = 0
    (x2 - 10x + 25) + (y2 - 2y + 1) = -3
    The right hand side is less than zero, so the equation has no graph.

    (f) 16x2 + 16y2 + 40x + 16y - 7 = 0
    Solution
    16x2 + 16y2 + 40x + 16y - 7 = 0
    x2 + y2 + 40⁄16x + y - 7⁄16 = 0
    x2 + 5⁄2x + 25⁄16 - 25⁄16 + y2 + y + 1⁄4 - 1⁄4 - 7⁄16 = 0
    (x2 + 5⁄2x + 25⁄16) - 25⁄16 + (y2 + y + 1⁄4) - 1⁄4 - 7⁄16 = 0
    (x2 + 5⁄2x + 25⁄16) - 25⁄16 + (y2 + y + 4⁄16) - 4⁄16 - 7⁄16 = 0
    (x + 5⁄4)2 + (y + 1⁄2)2 = 36⁄16 = 9⁄4
    (x + 5⁄4)2 + (y + 1⁄2)2 = (3⁄2)2
    The equation is a circle with center at (- 5⁄4, - 1⁄2) and radius is 3⁄2

    (g) x2 ⁄ 4 + y2 ⁄ 4 = 1
    Solution
    x2 ⁄ 4 + y2 ⁄ 4 = 1
    x2 + y2⁄4 = 1
    x2 + y2 = 4
    (x + 0)2 + (y + 0)2 = 22
    The graph is a circle with center at (0, 0) and radius is 2.

    (h) 2x2 + 2y2 + 8x + 7 = 0
    Solution
    2x2 + 2y2 + 8x + 7 = 0

    x2 + y2 + 4x + 7⁄2 = 0

    x2 + 4x + 4 - 4 + y2 + 7⁄2 = 0

    (x2 + 4x + 4) - 4 + y2 + 7⁄2 = 0

    (x2 + 4x + 4) + y2 - 4 + 7⁄2 = 0

    (x + 2)2 + (y + 0)2 = 1⁄2
    $$\small{ (x + 2)^2 + (y + 0)^2 = \left(\sqrt{\dfrac{1}{2}}\right)^2 = \left(\dfrac{1}{\sqrt{2}}\right)^2}$$ The graph is a circle with center at (-2, 0) and radius is 1⁄ √ 2