Lesson Identify elements of a circle given by its general equation

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Identify elements of a circle given by its general equation


This lesson shows you by examples how to identify elements of a circle given by its general equation.
The method is to transform the given general equation to the standard form using completing the squares.

Problem 1

Identify the center and the radius of a circle given by a general equation  x%5E2+%2B+y%5E2+-+6x+%2B+10y+-+15 = 0.

Solution

Apply completing the squares method to transform the given general equation of the circle to its standard form.

Make completing the squares separately for x-group terms and for y-group terms.


x%5E2+%2B+y%5E2+-+6x+%2B+10y+-+15 = 0   --->   (move the constant term to the right, if necessary)  --->

x%5E2+%2B+y%5E2+-+6x+%2B+10y = 15     --->   (group the terms containing x and y separately)  ---> 

%28x%5E2-6x%29+%2B+%28y%5E2%2B10y%29 = 15  --->   (Add constant terms to x-group and y-group to complete the squares)

%28x%5E2-2%2A%283%2Ax%29+%2B+9%29+%2B+%28y%5E2%2B2%2A%285%2Ay%29%2B25%29 = 15%2B9%2B25,   (add the same constant terms to the right side to keep the balance)  --->
%28x-3%29%5E2+%2B+%28y%2B5%29%5E2 = 49, %28x-3%29%5E2%2B+%28y%2B5%29%5E2 = 7%5E2. Thus you got the equation of the circle in the standard form.        The center of the circle is the point (3,-5). The radius of the circle is r = 7 units. Circle %28x-3%29%5E2 + %28y%2B5%29%5E2 = 49

Problem 2

Identify the center and the radius of a circle given by a general equation  2x%5E2+%2B+2y%5E2+-+8x+%2B+4y+-+14 = 0.

Solution

2x%5E2+%2B+2y%5E2+-+8x+%2B+4y+-+14 = 0  --->  (Divide both sides by 2 to get the leading coefficient at x%5E2 and y%5E2 equal to 1)  --->


Apply completing the squares method to transform the given general equation of the circle to its standard form.

Make completing the squares separately for x-group terms and for y-group terms.


x%5E2+%2B+y%5E2+-+4x+%2B+2y+-+7 = 0    --->   (move the constant term to the right, if necessary)  --->

x%5E2+%2B+y%5E2+-+4x+%2B+2y = 7      --->   (group the terms containing x and y separately)  ---> 

%28x%5E2-4x%29+%2B+%28y%5E2%2B2y%29 = 7   --->   (Add constant terms to x-group and y-group to complete the squares)

%28x%5E2-2%2A2x+%2B+4%29+%2B+%28y%5E2%2B2%2Ay%2B1%29 = 7%2B4%2B1,   (add the same constant terms to the right side to keep the balance)  --->
%28x-2%29%5E2+%2B+%28y%2B1%29%5E2 = 12, %28x-2%29%5E2%2B+%28y%2B1%29%5E2 = %28sqrt%2812%29%29%5E2. Thus you got the equation of the circle in the standard form.        The center of the circle is the point (2,-1). The radius of the circle is r = sqrt%2812%29 = 2%2Asqrt%283%29 units. Circle %28x-2%29%5E2 + %28y%2B1%29%5E2 = 12

My other lessons on ellipses in this site are
    - Ellipse definition, canonical equation, characteristic points and elements
    - Ellipse focal property
    - Tangent lines and normal vectors to a circle
    - Tangent lines and normal vectors to an ellipse
    - Optical property of an ellipse
    - Optical property of an ellipse revisited

    - Standard equation of an ellipse
    - Identify elements of an ellipse given by its standard equation
    - Find the standard equation of an ellipse given by its elements

    - General equation of an ellipse
    - Transform a general equation of an ellipse to the standard form by completing the square
    - Identify elements of an ellipse given by its general equation

    - Standard equation of a circle
    - Find the standard equation of a circle
    - General equation of a circle
    - Transform general equation of a circle to the standard form by completing the squares

    - OVERVIEW of lessons on ellipses

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.


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