Category Archives: 2.2 Circles

2.2 Circles, 2.2 Lines, 2.3 Lines, Intercpets

X-intercepts and Y-intercepts

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

In this picture, the graph crosses the x-axis at the ordered pair (2, 0).  Since every ordered pair on the x-axis has a y coordinate of zero we can let y=0 to find x-intercepts.

To find an x-intercept: Let y=0 and solve for x.

In this picture, the graph crosses the y-axis at the ordered pair (0, 6).  Since every ordered pair on the y-axis has a x coordinate of zero we can let x=0 to find y-intercepts.

To find an y-intercept: Let x=0 and solve for y.

2.2 Circles, 2.3 Circles, Analyze

Writing an equation of a circle in standard form

Example:  Write the equation of a circle in standard form.

x^2+y^2+4x-8y+11=0

Start by grouping the x terms together, grouping the y terms together and moving the constant to the other side of the equation.

x^2+4x+y^2-8y=-11

Use completing the square on the group of x terms and the group of y terms.

To find the number that completes the square for the x group, start with the coefficient of the x term, half it and square it.  The coefficient of the x term is 4

  (4/2)^2=4

To find the number that completes the square for the y group, start with the coefficient of the y term, half it and square it.  The coefficient of the y term is -8

  (-8/2)^2=16

Add these numbers to the group of x terms and the group of y terms.  Be careful to maintain the balance of the equation by adding the numbers to both sides of the equation.

x^2+4x+4+y^2-8y+16=-11+4+16

(x^2+4x+4)+(y^2-8y+16)=-11+4+16

Now the group of x terms is a perfect square trinomial and will factor to be a binomial squared.  The group of y terms will do the same.

(x+2)(x+2)+(y-4)(y-4)=9

(x+2)^2+(y-4)^2=9

The equation of the circle is written in standard form where it is easy to recognize the center and radius of the circle.

The center is (-2,4) and the radius is 9.

2.2 Circles, 2.3 Circles, Finding Intercepts of a Circle, Intercpets

Finding the Intercepts of a Circle Touching an Axis (Tangent to an axis)

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

To find an x-intercept: Let y=0 and solve for x.

To find an y-intercept: Let x=0 and solve for y.

 

Example:  Find the intercepts of the circle for the given equation.

(x-1)^2+(y+2)^2=4

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-1)^2+(y+2)^2=4

(x-1)^2+(0+2)^2=4

(x-1)^2+(2)^2=4

(x-1)^2+4=4

(x-1)^2+4-4=4-4

(x-1)^2=0

sqrt{(x-1)^2}=sqrt{0}

x-1=0

x-1+1=0+1

x=1

This equation has one x-intercept. (1,0)

To find a y-intercept, let x=0 and solve for y.

(x-1)^2+(y+2)^2=4

(0-1)^2+(y+2)^2=4

(-1)^2+(y+2)^2=4

1+(y+2)^2=4

(y+2)^2+1-1=4-1

(y+2)^2=3

sqrt{(y+2)^2}=sqrt{3}

y+2=pm sqrt{3}

y+2=pm sqrt{3}

y+2-2=-2 pm sqrt{3}

y=-2 pm sqrt{3}

Approximately y=-3.732 and x=-0.2679

This equation has two y-intercepts. (0,-2 + sqrt{3}) and (0,-2 - sqrt{3})

A tangent line to a circle may be defined as a line that intersects the circle in a single point.

This circle is tangent to the x-axis since it is touching the x-axis in a single point.  The x-axis (y=0) is the tangent line for the point on the circle (1,0).

Example:  Find the intercepts of the circle for the given equation.

(x-3)^2+(y-1)^2=9

Solution:

To find an x-intercept, let y=0 and solve for x.

(x-3)^2+(y-1)^2=9

(x-3)^2+(0-1)^2=9

(x-3)^2+(-1)^2=9

(x-3)^2+1=9

(x-3)^2+1-1=9-1

(x-3)^2=8

sqrt{(x-3)^2}=sqrt{8}

x-3=sqrt{4*2}

x-3=pm 2sqrt{2}

x-3+3=3 pm 2sqrt{2}

x=3 pm 2sqrt{2}

Approximately x=0.1716 and x=5.828

This equation has two x-intercepts. (3+2sqrt{2},0) and (3-2sqrt{2},0)

To find a y-intercept, let x=0 and solve for y.

(x-3)^2+(y-1)^2=9

(0-3)^2+(y-1)^2=9

(-3)^2+(y-1)^2=9

9+(y-1)^2=9

(y-1)^2+9-9=9-0

(y-1)^2=0

sqrt{(y-1)^2}=sqrt{0}

y-1=0

y-1+1=0+1

y=1

This equation has one y-intercept. (0,1).

This circle is tangent to the y-axis since it is touching the y-axis in a single point.  The y-axis (x=0) is the tangent line for the point on the circle (0,1).

2.2 Circles, 2.3 Circles, Finding Intercepts of a Circle, Intercpets

Finding the Intercepts of a Circle (4 Intercepts)

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

To find an x-intercept: Let y=0 and solve for x.

To find an y-intercept: Let x=0 and solve for y.

 

Example:  Find the intercepts of the circle for the given equation.

(x+3)^2+(y+6)^2=81

Solution:

To find an x-intercept, let y=0 and solve for x.

(x+3)^2+(y+6)^2=81

(x+3)^2+(0+6)^2=81

(x+3)^2+(6)^2=81

(x+3)^2+36=81

(x+3)^2+36-36=81-36

(x+3)^2=45

sqrt{(x+3)^2}=sqrt{45}

x+3=pm sqrt{9*5}

x+3=pm 3sqrt{5}

x+3-3=-3 pm 3sqrt{5}

x=-3 pm 3sqrt{5}

Approximately x=-9.708 and x=3.708

This equation has two x-intercepts. (-3 + 3sqrt{5},0) and (-3 - 3sqrt{5},0)

To find a y-intercept, let x=0 and solve for y.

(x+3)^2+(y+6)^2=81

(0+3)^2+(y+6)^2=81

(3)^2+(y+6)^2=81

9+(y+6)^2=81

(y+6)^2+9-9=81-9

(y+6)^2=72

sqrt{(y+6)^2}=sqrt{72}

y+6=pm sqrt{36*2}

y+6=pm 6sqrt{2}

y+6-6=-6 pm 6sqrt{2}

y=-6 pm 6sqrt{2}

Approximately y=-14.49 and x=2.485

This equation has two y-intercepts. (0,-6 + 6sqrt{2}) and (0,-6 - 6sqrt{2})

2.2 Circles, 2.3 Circles, Finding Intercepts of a Circle

Finding the Intercepts of a Circle (center at the origin)

An x-intercept is where the graph touches or crosses the x-axis.

A y-intercept is where the graph touches of crosses the y-axis.

To find an x-intercept: Let y=0 and solve for x.

To find an y-intercept: Let x=0 and solve for y.

 

Example:  Find the intercepts of the circle for the given equation.

x^2+y^2=25

Solution:

To find an x-intercept, let y=0 and solve for x.

x^2+y^2=25

x^2+(0)^2=25

x^2+0=25

x^2=25

sqrt{x^2}=sqrt{25}

x=pm 5

This equation has two x-intercepts. (5,0) and (-5,0)

To find a y-intercept, let x=0 and solve for y.

x^2+y^2=25

(0)^2+y^2=25

0+y^2=25

y^2=25

sqrt{y^2}=sqrt{25}

y=pm 5

This equation has two y-intercepts. (0,5) and (0,-5)

 

2.2 Circles, 2.3 Circles, Find the Equation

Write the Equation of a Circle Given the center and the radius

Example:  Write the equation of a circle in standard form given the center of the circle is (5,7) and the radius of the circle is 6.  Then write the equation in general (expanded) form.

Solution:

The standard form of an equation of a circle is (x-h)^2+(y-k)^2=r^2 where r is the radius and   (h, k) is the center.

The center of our circle is (5, 7).  So h=5 and k=7.

The radius of our circle is 6 so r=6.

Replace h, k and r in standard form of an equation of a circle.

(x-5)^2+(y-7)^2=6^2

Simplify.

(x-5)^2+(y-7)^2=36

The equation of a circle in standard for with center (5,7) and a radius of 6 is (x-5)^2+(y-7)^2=36.

To write the equation in general form we can start with the standard form we just found and multiply each binomial.

(x-5)^2+(y-7)^2=36

(x-5)(x-5)+(y-7)(y-7)=36

x^2-5x-5x+25+y^2-7y-7y+49=36

x^2-10x+25+y^2-14y+49=36

x^2-10x+25+y^2-14y+49-36=36-36

x^2-10x+y^2-14y+38=0

x^2+y^2-10x-14y+38=0

The equation of a circle in general form with center (5,7) and a radius of 6 is x^2+y^2-10x-14y+38=0.

2.2 Circles, 2.3 Circles, Analyze

Find the center and the radius from the equation

A circle is the collection of points that are equidistant to a center point.  The distance is the radius denoted r.  The center is denoted (h, k).

The standard form of an equation of a circle is (x-h)^2+(y-k)^2=r^2 where r is the radius and   (h, k) is the center.

 

Example:  Find the center and the radius from the given equation.

(x-2)^2+(y-3)^2=16

Solution:  If you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=2.  In the binomial with y, the number after the minus sign is k.  Thus, k=3.  The r^2 lines up with 16.  So r^2=16 which means r=4.

This is the equation of a circle with a center of (2, 3) and a radius of r=4 .

 

Example:  Find the center and the radius from the given equation.

(x+1)^2+(y-4)^2=25

Solution:  Notice that one of the binomials has a + instead of the – that is in the standard form of the circle.  Rewrite the addition as subtraction.

x+1 is the same as x-(-1).

The rewritten equation is as follows.

(x-(-1))^2+(y-4)^2=25

Then, if you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=-1.  In the binomial with y, the number after the minus sign is k.  Thus, k=4.  The r^2 lines up with 25.  So r^2=25 which means r=5.

This is the equation of a circle with a center of (-1, 4) and a radius of r=5 .

 

Example:  Find the center and the radius from the given equation.

x^2+(y+7)^2=12

Solution:  Notice that one of the binomials has a + instead of the – that is in the standard form of the circle.  Rewrite the addition as subtraction.

y+7 is the same as y-(-7).

The other binomial doesn’t have any number added or subtracted.  We can rewrite this by subtracting zero.

  x is the same as x-0.

The rewritten equation is as follows.

(x-0)^2+(y-(-7))^2=12

Then, if you line up the standard form of a circle with the equation given you can determine the center and the radius.

In the binomial with x, the number after the minus sign is h.  Thus, h=0.  In the binomial with y, the number after the minus sign is k.  Thus, k=-7.  The r^2 lines up with 25.  So r^2=12 which means r=sqrt{12}=2sqrt{3}.

This is the equation of a circle with a center of (0, -7) and a radius of r=2sqrt{3} .