Category Archives: MAC1140

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)

 

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.

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} .