This lesson explains how to obtain and use the standard form of
the equation of a circle in Geometry.
Further you’ll also explore how to convert between the standard and general form of the
circle
equation. You can take a look at the circles and learn discuss how to write an equation in
standard form given either the radius and center or the equation written in general form.
As the next step, can take help on using the idea of completing the square to go from the
general form to the standard form of the equation of the circle. Included are the relevant
problems finding the solutions to find the center and radius of a circle looking at the
equation of circle? The contents will be presented by the instructor in own handwriting
with the help of several examples with solutions and using watch video.
(Continued from above)
The standard form of a circle is (x  h)^{2}
+ (y  k)^{2} = r^{2},
where (h, k) is the center
and r is the radius of a circle. Also, when the center of the circle is (0, 0), and
the radius is r, then the equation of a circle is x^{2} + y^{2}
= r^{2}. Thus if the radius of a
circle is 2 with the center at origin, the equation of the circle is
x^{2}
+ y^{2} = 4. Every circle centered at the origin will have this type of equation.
Now take a case to know the steps for finding the coordinates of the
center and the radius of a circle from equation of a circle. E.g. find
the center and radius of the circle whose equation is (x  5)^{2} + (y  1)^{2}
= 36. Look a this equation in line with its standard form like:
(x  h)^{2} + (y
 k)^{2} = r^{2}
and you’ll we get the result, say h = 5, k = 1
and r =
36= 6. Thus, center of the circle is (5, 1) and
r = 6. Another
illustration, if the center of circle is (3, 1) and radius 7; to find
the equation of the circle you plug in these values in the equation in
standard form (x  h)^{2}
+ (y  k)^{2} = r^{2}
and it results in the equation i.e. (x  3)^{2}
+ (y + 1)^{2} = 49.
For example: Given write the equation x^{2} + y^{2}
 4x + 6y +4 =0 into the
standard form for the circle, and find its radius.
You can follow the steps given below:
The standard form of a circle is (x  h)^{2}
+ (y  b)^{2}= r^{2},
where (h, k) is the center and r is the radius of a circle.
Notice here the term x^{2}  4x; and half of 4 is 2. Then (2)^{2} = 4
Next pick up x^{2}  4x + 4 and factor, that results into (x  2)^{2}.
Now you are left with y^{2}+ 6y. Then
6 / 3 = 2, and
(3)^{2} = 9.
Add 9 to both the sides, and that results into (x  2)^{2}
+ y^{2} + 6y + 9 = 9.
Note that y^{2}+ 6y + 9 = (y + 3)^{2}
and 9 = 3^{2},
so the equation is (x  2)^{2}
+ (y + 3)^{2} = 3^{2}.
Identify the circle now: It is the circle with center at (2, 3) and radius 3.
Further if you want to draw this circle, use a compass. On the graph sheet,
mark the center and use compass to draw a circle with 3 unit radius.
You’ll find some more examples now to develop better understanding that should help to
solve the problems:
Example 1: Look at another example (Figure on right Not to Scale): The circle
is located at (1, 2) and having radius 1.5 unit. When you plug in these
values in the equation in standard form
(x  h)^{2}
+ (y  k)^{2} = r^{2},
it results in the equation (x + 1)^{2} +
(y + 3)^{2} = (1.5)^{2}
= 2.25. Important
to note here: the direction of translation is opposite to as appears in
the equation, the reason being that the shift is with the variable.
Example 2: Now look at a different case where equation of the circle is not in standard form. E.g.
Given: Find the center and radius of the circle, x^{2}
+ y^{2} + 6y + 8 = 0, and graph it.
Step 1: Write the equation of the circle in standard form
(x  h)^{2}
+ (y  k)^{2} = r^{2}
.
It can be noticed that the coefficients of x^{2}
and y^{2} in the given equation are equal and with the same sign, it is obvious that this equation is circle. If it had different numbers it would be ellipse, and with opposite signs it would be the case of hyperbola (by virtue of known properties).
Step 2: Isolating the x^{2}, x, y^{2},
and y terms results into x^{2} + y^{2}
+ 6y = 8
Step 3: Complete the square for x and y. Since x is the missing term here, its coefficient is taken to be ‘0’. Half of 0 is zero, and also its square is also zero. Similarly complete the square for y i.e. half of 6 is 3 and when squared it results to 9.
Step 4: Factoring as binomial squared gives, x^{2}
+ (y + 3)^{2} = 1
Step 5: Compare this with standard form, and identify center as well as radius. You’ll find h = 0,
k = 3, and r = 1.
Thus the center of circle is (0, 3) and the radius is 1, as the final answer. You can draw the graph as shown in Example 1, above.
Example 3: Given the circle equation is,
(x + 3)^{2} + (y  4)^{2}
= 16. Find out, if the
point A (4, 5) is inside, outside or on the circle.
Step 1: Find out the distance from the center of the circle to point A.
In the given equation,
center O at (3 , 4)
radius r = (16) = 4
Step 2: Distance between O and A = {[4  (3)]^{2}
+ [5  4]^{2}} = (7^{2} +1^{2}) =
(50)
Since the distance from O to A is (50)^{2}
= 7.07, which is greater
than radius 4, point A is outside the circle. You can verify it graphically also.
Remember: Circles with centers different than the origin reflect horizontal
and/or vertical translations in their equations. Also that the values for the
translation(s) are including the respective variables, the shifts are always
in 'opposite' direction than appearing in the equation.
The video above will explain more in detail about the Equation of a circle in
standard form, with the help of several examples.
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