Sullivan Chapter 2 | math15fun.com

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 .

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.

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} .