Circumference equation – Circumference — Wikipedia

Equation of the circumference II: general equation

A circumference with center C = (a, b) and radius r can be rewritten in light of the reduced equation as:

Developing the squares of the above mentioned equation we obtain:

x^2+y^2-2ax-2by+a^2+b^2-r^2=0

and doing the change A= -2a, \ \ B=-2b, \ \ C=a^2+b^2-r^2 in:

x^2+y^2-2ax-2by+a^2+b^2-r^2=0

the new equation is obtained:

This way we have found another analytical expression that defines the points of a circumference. This is the general equation of the circumference.

Let’s see how to determine the radius and the center of a circumference from the general equation.

We can do the following:

A=-2a, \ \ B=-2b, \ \ C=a^2+b^2-r^2

We isolate these expressions in terms of a, b and r. We have:

\displaystyle a=-\frac{A}{2} r^2=a^2+b^2-C=\Big(-\frac{A}{2}\Big)^2+\Big(-\frac{B}{2}\Big)^2-C=\frac{A^2+B^2-4C}{4}

And since we know that, in the limited expression, (a, b) is the center and r the radius, given a general equation:

the center of such a circumference is the point \displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big) and the radius is \displaystyle r=\sqrt{\frac{A^2+B^2-4C}{4}}.

Let’s suppose that they give us the circumference:

then we see that it is centred at the point:
\displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big)=\Big(-\frac{-2}{2},-\frac{4}{2}\Big)=(1,-2)

\displaystyle r=\sqrt{\frac{A^2+B^2-4C}{4}}=\sqrt{\frac{(-2)^2+4^2-4\cdot(- 4)}{4}}=
=\displaystyle\sqrt{\frac{4+16+16}{4}}=\sqrt{\frac{36}{4}}=\frac{6}{2}=3

Let’s now see the inverse process, that is to say:

Giving the general equation of the circumference that has, for example, radius 4 and center (-5, 6).

We write the reduced equation:

(x-a)^2+(y-b)^2=r^2 \Rightarrow (x+5)^2+(y-6)^2=4^2

developing the squares we have:

(x+5)^2+(y-6)^2=4^2 \Rightarrow x^2+10x+25+y^2-12y+36=16

If we rearrange it and add all the independent terms, we obtain the general equation of the above mentioned circumference, which is:

Let’s see what happens when the circumference is centred on the origin and we want to write its general equation:

Since (0, 0) is the center we have: a=0 and b=0 for which reason,

\left.{\begin{matrix} {0=a=-\frac{A}{2}} \\ {0=b=-\frac{B}{2}} \end{matrix}}\right \}\Longrightarrow{\left \{ {\begin{matrix} {A=0}\\{B=0}\end{matrix}}\right . }

So that in the general equation, only quadratic terms and independent terms will exist, that is to say:

Moving the constant term to the other side we obtain:

where we know that:

since the supposed center was (0, 0).

Note that for the circunference centered at the origin both equations are very similar.

Let’s see an example:

Circumference centred on the origin and radius 7.

Reduced equation: x^2+y^2=7^2

General equation: x^2+y^2+C=0 where C=-7^2 \Longrightarrow x^2+y^2-7^2=0

Summing up we have:

Considering the circumference: (x-a)^2+(y-b)^2=r^2

Then the center is the point of the plane (a,b) and the radius is r.

(x-8)^2+(y+3)^2=1 has center at (8,-3) and radius 1.

Considering the circumference: x^2+y^2+Ax+By+C=0

Then the center is at the point of the plane \displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big) and the radius is \displaystyle r=\sqrt{\Big(\frac{A}{2}\Big)^2+\Big(\frac{B}{2}\Big)^2-C}

x^2+y^2+x-5y-2=0 has center \displaystyle \Big(\frac{-1}{2},\frac{5}{2}\Big) and radius

\displaystyle r=\sqrt{\Big(\frac{1}{2}\Big)^2+\Big(\frac{-5}{2}\Big)^2-(-2)}=\sqrt{\frac{1+25+8}{4}}=\sqrt{\frac{34}{4}}=\sqrt{\frac{17}{2}}

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Circumference of a Circle Equation

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07-Mar-2016

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To simplify calculations the value of is rounded to 3.14. The diameter of a circle is twice its radius. Therefore, Circumference, C = 2 r The distance around a closed curve is called as the circumference of the circle. It is also defined as the length around the circle. The circumference of a circle is measured in linear units like inches or centimeters. Know More About What is the Volume of a Sphere Circumference of a Circle Examples Circumference of a Circle Formula

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Circumference of a Circle Equation

• Published on
05-Mar-2016

• View
214

2

DESCRIPTION

The distance around a closed curve is called as the circumference of the circle. It is also defined as the length around the circle. The circumference of a circle is measured in linear units like inches or centimeters. Know More About Geometry Distance Formula Circumference of a Circle Formula

Transcript

Circumference of a Circle EquationKnow More About Geometry Distance FormulaThe distance around a closed curve is called as the circumference of the circle. It is also defined as the length around the circle. The circumference of a circle is measured in linear units like inches or centimeters.Circumference of a Circle FormulaThe formula of circumference of a circle is given by the following formula,Circumference, C = 2 * * rwhere, r is the radius of the circle. and pi is a constant value and is equal to 3.14The circumference of a circle can be calculated by its diameter using the following formula:Circumference, C = * DWhere, value of is a constant and its value is approximately 3.14159265358979323846…. or 22/7 to be more precise and D is the diameter of the circle. To simplify calculations the value of is rounded to 3.14. The diameter of a circle is twice its radius. Therefore, Circumference, C = 2 r Circumference of a Circle ExamplesBelow are the examples on circumference of a circle -Example 1:Find the circumference of a circle given that its radius is 10.Solution:The circumference is always multiply by 2 the length of the radius,FormulaCircumference of circle = 2 rCircumference = 2 x 10 x = 62.8Example 2:Find the circumference of a circle known that area of the circle is 153.86Solution:Step 1:Learn More What is Distance Formula Learn about distance formula here and understand the concept better with solved examples provided. Students can also use the online distance formula calculator and distance formula worksheet provided in the page.Let’s understand what is the distance formula? The length of a line segment AB, which joins A (x1, y1) and B (x2, y2) is given by,Distance Formula ProofLet A (x1, y1) and B (x2, y2) be two points in the plane.Let d = distance between the points A and B.Draw AL and BM perpendicular to x-axis (parallel to y-axis).Draw AC perpendicular to BM to cut BM at C.In the figure,What is Distance Formula OL = x1, OM = x2 [AC = LM = OM — OL = x2 — x1]MB = y2, MC = LA = y1 [CB = MB — MC = y2 — y1]From the right-angled DACB,Distance Formula ExamplesBelow are some examples based on distance formulaExample 1: Find the distance between the following pair of points: A (1,2) and B (4,5).Solution:Using the distance formula, we haveExample 2: Find the distance between places when the two coordinates (2, 4) and (4, 6)are given, using the distance formula.?Solution:(x1, y1)= (2, 4)(x2, y2) = (4, 6)Read More About Pyramid Definition ThankYouTutorVista.comSlide 1Slide 2Slide 3Slide 4Slide 5

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