How To Find The Equation Of A Circle

Equation of Circle

The equation of circle provides an algebraic manner to draw a circumvolve, given the middle and the length of the radius of a circle. The equation of a circumvolve is unlike from the formulas that are used to calculate the area or the circumference of a circle. This equation is used across many problems of circles in coordinate geometry.

To represent a circumvolve on the Cartesian plane, nosotros require the equation of the circumvolve. A circle can exist drawn on a piece of newspaper if we know its heart and the length of its radius. Similarly, on a Cartesian plane, we can describe a circle if we know the coordinates of the heart and its radius. A circle can exist represented in many forms:

• General form
• Standard form
• Parametric form
• Polar form

In this commodity, let's larn about the equation of the circle, its various forms with graphs and solved examples.

ane.	What is the Equation of Circle?
2.	Dissimilar Forms of Equation of Circumvolve
3.	Equation of a Circle Formula
4.	Derivation of Circle Equation
5.	Graphing the Equation of Circumvolve
6.	How to Find Equation of Circle?
7.	Converting General Form to Standard Form
eight.	Converting Standard Form to General Form
nine.	FAQs on Equation of a Circle

What is the Equation of Circumvolve?

An equation of a circumvolve represents the position of a circle in a Cartesian plane. If we know the coordinates of the center of the circumvolve and the length of its radius, we tin can write the equation of a circumvolve. The equation of circle represents all the points that lie on the circumference of the circle.

A circle represents the locus of points whose altitude from a stock-still point is a constant value. This fixed point is called the center of the circumvolve and the constant value is the radius r of the circumvolve. The standard equation of a circle with center at \((x_1, y_1)\) and radius r is \( (x - x_1)^two + (y - y_1)^2 = r^2\).

Unlike Forms of Equation of Circle

An equation of circumvolve represents the position of a circumvolve on a cartesian plane. A circle can be fatigued on a piece of paper given its middle and the length of its radius. Using the equation of circumvolve, one time nosotros discover the coordinates of the center of the circle and its radius, we will be able to draw the circle on the cartesian aeroplane. There are dissimilar forms to represent the equation of a circle,

• General form
• Standard form
• Parametric grade
• Polar grade

Let'south look at the two mutual forms of the equation of circle-general grade and standard grade of the equation of circle hither along with the polar and parametric forms in particular.

General Equation of a Circumvolve

The general form of the equation of a circumvolve is: x² + y^two + 2gx + 2fy + c = 0. This full general course is used to find the coordinates of the center of the circle and the radius, where m, f, c are constants. Dissimilar the standard form which is easier to empathise, the full general form of the equation of a circumvolve makes information technology difficult to find any meaningful properties about any given circle. And then, we will be using the completing the foursquare formula to brand a quick conversion from the full general form to the standard grade.

Standard Equation of a Circle

The standard equation of a circumvolve gives precise information about the heart of the circumvolve and its radius and therefore, it is much easier to read the center and the radius of the circle at a glance. The standard equation of a circle with center at \((x_1, y_1)\) and radius r is \( (x - x_1)^2 + (y - y_1)^2 = r^2\), where (x, y) is an arbitrary point on the circumference of the circumvolve.

The distance between this indicate and the middle is equal to the radius of the circle. Allow's apply the distance formula betwixt these points.

\(\sqrt{(ten - x_1)^two + (y - y_1)^two} = r\)

Squaring both sides, we get the standard form of the equation of the circle as:

\((x - x_1)^ii + (y - y_1)^2 = r^2\)

Consider this case of an equation of circle (x - four)² + (y - 2)² = 36 is a circle centered at (4,ii) with a radius of 6.

Parametric Equation of a Circle

We know that the general form of the equation of a circle is x² + y² + 2hx + 2ky + C = 0. Nosotros take a general indicate on the purlieus of the circle, say (x, y). The line joining this general indicate and the eye of the circle (-h, -one thousand) makes an angle of \(\theta\). The parametric equation of circle can be written as x² + yⁱⁱ + 2hx + 2ky + C = 0 where x = -h + rcosθ and y = -k + rsinθ.

Polar Equation of a Circle

The polar grade of the equation of the circumvolve is well-nigh similar to the parametric course of the equation of circumvolve. We usually write the polar form of the equation of circle for the circumvolve centered at the origin. Permit's take a betoken P(rcosθ, rsinθ) on the boundary of the circumvolve, where r is the altitude of the betoken from the origin. We know that the equation of circumvolve centered at the origin and having radius 'p' is xⁱⁱ + y² = p².

Substitute the value of x = rcosθ and y = rsinθ in the equation of circle.

(rcosθ)² + (rsinθ)ⁱⁱ = pⁱⁱ
rⁱⁱcos²θ + r²sinⁱⁱθ = p²
r²(cos^twoθ + sin^twoθ) = pⁱⁱ
r²(1) = p²
r = p
where p is the radius of the circle.

Example: Find the equation of the circumvolve in the polar course provided that the equation of the circumvolve in standard form is: x^two + y² = 9.

Solution:

To find the equation of the circle in polar class, substitute the values of \(x\) and \(y\) with:

x = rcosθ
y = rsinθ

x = rcosθ
y = rsinθ
x² + y^two = ix
(rcosθ)² + (rsinθ)² = nine
r^twocos²θ + r²sin²θ = 9
r²(cos^twoθ + sin²θ) = nine
r²(one) = 9
r = 3

Equation of a Circle Formula

The equation of a circle formula is used for calculating the equation of a circle. Nosotros can find the equation of whatsoever circle, given the coordinates of the center and the radius of the circle past applying the equation of circumvolve formula. The equation of circle formula is given as, \((x - x_1)^2 + (y - y_1)^two = r^2\).

where,\((x_1, y_1)\) is the center of the circle with radius r and (x, y) is an capricious point on the circumference of the circle.

Derivation of Circle Equation

Given that \((x_1, y_1)\) is the center of the circle with radius r and (ten, y) is an arbitrary point on the circumference of the circle. The altitude betwixt this point and the middle is equal to the radius of the circumvolve. And then, let'southward apply the distance formula between these points.

\(\sqrt{(x - x_1)^2 + (y - y_1)^2} = r\).

Squaring both sides, nosotros get: \((x - x_1)^ii + (y - y_1)^2 = r^2\). So, the equation of a circle is given by:

\((x - x_1)^2 + (y - y_1)^2 = r^2\)

Case: Using the equation of circumvolve formula, discover the center and radius of the circumvolve whose equation is (x - i)ⁱⁱ + (y + ii)ⁱⁱ = 9.

Solution:

We will use the circle equation to determine the centre and radius of the circle.
Comparison \((x - i)^2 + (y + 2)^2 = ix\) with \((10 - x_1)^ii + (y - y_1)^ii = r^2\), we become

\(x_1\) = 1, \(y_1\) = -2 and r = iii

And then, the center and radius are (1, -ii) and three respectively.

Respond: The center of the circle is (1, -2) and its radius is 3.

Graphing the Equation of Circumvolve

In order to show how the equation of circumvolve works, let'south graph the circumvolve with the equation (10 -3)^two + (y - ii)² = ix. At present, earlier graphing this equation, we need to brand sure that the given equation matches the standard form \((ten - x_1)^ii + (y - y_1)^2 = r^2\).

• For this, we just need to change the constant 9 to match with r² equally (x -three)^two + (y - 2)² = 3^two.
• Here, we need to note that one of the common mistakes to commit is to consider \(x_{ane}\) as -three and \(y_{1}\) equally -2.
• In the equation of circle, if the sign preceding \(x_{one}\) and \(y_{one}\) are negative, then \(x_{1}\) and \(y_{1}\) are positive values and vice versa.
• Hither, \(x_{i}\) = 3, \(y_{1}\) = 2 and r = 3

Thus, the circle represented past the equation (10 -3)ⁱⁱ + (y - 2)^two = 3², has its centre at (three, 2) and has a radius of three. The below-given image shows the graph obtained from this equation of the circle.

How to Find Equation of Circle?

There are so many different ways of representing the equation of circle depending on the position of the circle on the cartesian plane. We have studied the forms to correspond the equation of circle for given coordinates of center of a circumvolve. There are certain special cases based on the position of the circle in the coordinate plane. Let's acquire about the method to find the equation of circle for the full general and these special cases.

Equation of Circle With Heart at (ten\(_1\), y\(_1\))

To write the equation of circle with center at (x\(_1\), y\(_1\)), we will use the post-obit steps,

• Step 1: Note down the coordinates of the center of the circumvolve(x\(_1\), y\(_1\)) and the radius 'r'.
• Step 2: Apply the equation of circle formula, \(\sqrt{(x - x_1)^2 + (y - y_1)^2} = r\).
• Stride 3: Express the respond in the required circle equation form.

Equation of Circle With Middle at the Origin

The simplest case is where the circle'south center is at the origin (0, 0), whose radius is r. (x, y) is an arbitrary point on the circumference of the circumvolve.

The distance between this point and the center is equal to the radius of the circumvolve. Permit'southward use the altitude formula between these points.

\( \sqrt{(10 - 0)^2 + (y - 0)^2} = r\)

Squaring both sides, we get:

\( (x - 0)^two + (y - 0)^ii = r^2\)

\( x^ii + y^2 = r^ii\)

Example: What volition be the equation of a circle if its center is at the origin?

Solution:

The equation of a circumvolve is given past \((ten - x_1)^2 + (y - y_1)^ii = r^ii\).

If middle is at origin, then \(x_1\)= 0 and \(y_1\)= 0.

Answer: The equation of the circle if its center is at origin is x²+ y²= r².

Equation of Circle With Eye on x-Axis

Consider the example where the centre of the circle is on the 10-axis: (a, 0) is the center of the circle with radius r. (ten, y) is an capricious point on the circumference of the circle.

The altitude between this point and the center is equal to the radius of the circle. Let's employ the distance formula between these points.

\(\sqrt{(ten - a)^2 + (y - 0)^2} = r\)

Squaring both sides, we go:

\((x - a)^2 + (y - 0)^two = r^two\)

\((x - a)^two + (y)^ii = r^2\)

Equation of Circle With Centre on Y-Centrality

Consider the example where the center of the circle is on the y-centrality: (0, b) is the heart of the circumvolve with radius r. (x, y) is an arbitrary betoken on the circumference of the circle.

The altitude between this betoken and the middle is equal to the radius of the circle. Permit's utilize the distance formula between these points.

\( \sqrt{(x - 0)^ii + (y - b)^2} = r\)

Squaring both sides, nosotros get:

\( (x - 0)^2 + (y - b)^ii = r^2\)

\( (x)^2 + (y - b)^2 = r^2\)

Equation of Circumvolve Touching ten-Axis

Consider the case where the circumference of the circumvolve is touching the ten-centrality at some point: (a, r) is the center of the circle with radius r. If a circumvolve touches the x-axis, then the y-coordinate of the center of the circle is equal to the radius r.

(x, y) is an capricious bespeak on the circumference of the circumvolve. The distance between this bespeak and the eye is equal to the radius of the circle. Let'due south apply the altitude formula between these points.

\( \sqrt{(ten - a)^2 + (y - r)^two} = r\)

Squaring both sides, nosotros get:

\( (10 - a)^ii + (y - r)^2 = r^2\)

Equation of Circle Touching y-Centrality

Consider the case where the circumference of the circle is touching the y-axis at some bespeak: (r, b) is the center of the circumvolve with radius r. If a circumvolve touches the y-axis, then the x-coordinate of the center of the circle is equal to the radius r.

(ten, y) is an arbitrary point on the circumference of the circle. The altitude between this point and the center is equal to the radius of the circle. Permit's apply the distance formula betwixt these points.

\(\sqrt{(10 - r)^2 + (y - b)^2} = r\)

Squaring both sides

\((x - r)^2 + (y - b)^2 = r^2\)

Equation of Circle Which Touches Both the Axes

Consider the instance where the circumference of the circumvolve is touching both the axes at some point: (r, r) is the centre of the circumvolve with radius r. If a circle touches both the x-centrality and y-axis, then both the coordinates of the center of the circumvolve get equal to the radius (r, r).

(x, y) is an arbitrary point on the circumference of the circumvolve. The altitude between this indicate and the centre is equal to the radius of the circle. Let'south use the distance formula between these points.

\(\sqrt{(x - r)^2 + (y - r)^2} = r\)

Squaring both sides

\((ten - r)^ii + (y - r)^2 = r^ii\)

If a circumvolve touches both the axes, and then consider the middle of the circle to be (r,r), where r is the radius of the circle. Here, (r,r) tin can be positive as well as negative. For example, the radius of the circle is 3 and information technology is touching both the axes, then the coordinates of the centre can exist (three,3), (three,−three), (−3,iii), or (−3,−3).

Instance: If the equation of circle in general course is given as \(10^2 + y^2 + 6x + 8y + 9 = 0\), find the coordinates of the heart and the radius of the circle.

Solution:

Given the equation of the circle \( ten^2 + y^two +6x + 8y + ix = 0\)

The general form of the equation of the circle with centre \((x_1, y_1)\) and radius \(r\) is \( x^2 + y^ii + Ax + By + C = 0\)
where \(A = -2x_1\)
\(B = -2y_1\)
\(C = {x_1}^ii + {y_1}^2 -r^2\)

From the equation of the circle \( 10^2 + y^2 +6x + 8y + 9 = 0\)

\(A = 6 \\
-2x_1 = half-dozen \\
x_1 = -3 \\
B = 8 \\
-2y_1 = eight \\
y_1 = -4 \\
C = ix \\
{x_1}^two + {y_1}^2 -r^2 = nine \\
{-3}^2 + {-4}^two -r^2 = ix \\
9 + 16 -r^two = 9 \\
r^2 = 16 \\
r = 4 \)

Converting General Form to Standard Form

This is the standard equation of circle, with radius r and middle at (a,b): (x - a)^two + (y - b)ⁱⁱ = r^two and consider the general grade as: x² + yⁱⁱ + 2gx + 2fy + c = 0. Here are the steps to be followed to convert the full general course to the standard form:

Footstep 1: Combine the like terms and take the abiding on the other side as 10² + 2gx + y² + 2fy = - c -> (one)

Step ii: Use the perfect square identity (x + g)² = x^two + 2gx + g² to find the values of the expression 10² + 2gx and y² + 2fy as:

(x + g)² = x² + 2gx + one thousand² ⇒ x² + 2gx = (x + g)² - g² -> (2)

(y + f)² = y² + 2fy + f² ⇒ y^two + 2fy = (y + f)^two - f² -> (3)

Substituting (2) and (three) in (ane), we get the equation as:

(ten+grand)^two - gⁱⁱ + (y+f)² - f^two = - c

(x+grand)² + (y+f)² = g^two + f² - c

Comparing this equation with the standard form: (10 - a)² + (y - b)² = r² we get,

Center = (-g,-f) and radius = \(\sqrt{1000^2+f^ii - c}\)

We need to make sure that the coefficients of x^two and yⁱⁱ are i before applying the formula.

Consider an example here to notice the middle and radius of the circle from the general equation of the circle: x² + y² - 6x - 8y + 9 = 0.

The coordinates of the centre of the circle can exist constitute equally: (-m,-f). Here g = -half dozen/2 = -3 and f = -8/ii = -4. So, the heart is (3,4).

Radius r = \(\sqrt{g^2+f^2 - c}\) = \(\sqrt{(-3)^{2}+(-iv)^{ii} - nine}\) = \(\sqrt{9 + 16 - 9}\) = \(\sqrt{16}\) = 4. So, radius r = 4.

Converting Standard Form to General Form

We tin can utilise the algebraic identity formula of (a - b)² = a² + b² - 2ab to convert the standard form of equation of circumvolve into the general form. Allow'due south run across how to exercise this conversion. For this, expand the standard grade of the equation of the circle as shown below, using the algebraic identities for squares:

\( (10 - x_1)^two + (y - y_1)^2 = r^2\)

\( 10^two +{x_1}^2 -2xx_1 + y^2 +{y_1}^ii -2yy_1 = r^ii\)
\( x^two + y^2 - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^2 = r^2\)
\( x^2 + y^ii - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^ii -r^2 = 0\)

Replace \(-2x_1\) with 2g, \(-2y_1\) with 2f, \( {x_1}^2 + {y_1}^two -r^two\) with \(c\), we go:

\( x^2 + y^2 + 2gx + 2fy + c = 0\)

At present, we get the general form of equation of circle as: \( ten^two + y^2 + 2gx + 2fy + c = 0\), where g, f, c are constants.

Related Articles on Equation of Circle

Check out the following pages related to the equation of circumvolve

• Equation of a Circle Calculator
• Circumference of a Circle
• All Circle Formulas
• Ratio of Circumference to Bore

Of import Notes on Equation of Circle

Here is a list of a few points that should exist remembered while studying the equation of circle

• The full general form of the equation of circle e'er has x² + y² in the beginning.
• If a circle crosses both the axes, then there are four points of intersection of the circle and the axes.
• If a circle touches both the axes, then in that location are only two points of contact.
• If any equation is of the class \(10^ii + y^2 + axy + C = 0\), then it is not the equation of the circle. There is no \(xy\) term in the equation of circle.
• In polar form, the equation of circle always represents in the form of \(r\) and \(\theta\).
• Radius is the altitude from the center to any point on the purlieus of the circle. Hence, the value of the radius of the circle is always positive.

Examples on Circumvolve Equations

1. Example 1: Find the equation of the circle in standard class for a circle with center (two,-3) and radius 3.

   Solution:

   Equation for a circle in standard form is written every bit: (x - x\(_1\))² + (y - y\(_1\))^two = r². Here, (x\(_1\), y\(_1\)) = (2, -3) is the center of the circle and radius r = 3.

   Let'due south put these values in the standard form of equation of circumvolve:

   (x - two)^two + (y - (-iii))ⁱⁱ = (3)²
   (x - 2)^two + (y + three)² = nine is the required standard course of the equation of the given circle.

2. Example 2: Write the equation of circle in standard form for a circle with center (-1, 2) and radius equal to vii.

   Solution:

   Equation for a circle in standard class is written as: (x - 10\(_1\))² + (y - y\(_1\))² = r². Here (ten\(_1\), y\(_1\)) = (-1, ii) is the center of the circle and radius r = 7.

   Let's put these values in the standard course of equation of circle:

   (ten - (-1))² + (y - 2)² = 7^two
   (10 + ane)^two + (y - 2)² = 49 is the required standard form of the equation of the given circumvolve.

3. Example 3: Find the equation of the circle in the polar form provided that the equation of the circle in standard form is: x^two + y² = sixteen.

   Solution:

   To discover the equation of the circle in polar form, substitute the values of x and y with:

   x = rcosθ
   y = rsinθ

   xⁱⁱ + y² = xvi

   (rcosθ)^two + (rsinθ)ⁱⁱ = xvi

   rⁱⁱcos²θ + r^twosinⁱⁱθ = 16

   r²(1) = iv

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Practice Questions on Equation of Circle

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FAQs on Equation of Circle

What is the Equation of Circumvolve in Geometry?

The equation of circle represents the locus of point whose distance from a fixed indicate is a constant value. This stock-still point is called the center of the circumvolve and the constant value is the radius of the circle. The standard equation of circle with middle at \((x_1, y_1)\) and radius r is \( (x - x_1)^2 + (y - y_1)^2 = r^two\).

What is the Equation of Circle When the Center Is at the Origin?

The simplest example is where the circle's center is at the origin (0, 0), whose radius is r. (x, y) is an arbitrary bespeak on the circumference of the circle. The equation of circle when the center is at the origin is xⁱⁱ + y^two = r^two.

What is the General Equation of Circumvolve?

The full general class of the equation of circle is: x² + y² + 2gx + 2fy + c = 0. This general form of the equation of circle has a centre of (-yard, -f), and the radius of the circle is r = \(\sqrt{g^two + f^two - c}\).

What is the Parametric Equation of Circle?

The parametric equation of circumvolve can be written as \(x^2 + y^2 + 2hx + 2ky + C = 0\) where \(x = -h +rcos \theta\) and \(y = -chiliad +rsin \theta\)

What is C in the Full general Equation of Circumvolve?

The full general form of the equation of circle is: x² + y² + 2gx + 2fy + c = 0. This general form is used to notice the coordinates of the heart of the circle and the radius of the circle. Hither, c is a abiding term, and the equation having c value represents a circle that is not passing through the origin.

What are the Various Forms of Equations of a Circumvolve?

Let'south expect at the two common forms of the equation are:

• General Form x^two + y^two + 2gx + 2fy + C = 0
• Standard Grade \((x - x_1)^two + (y - y_1)^2 = r^2\)

What is the Equation of Circle When the Middle is on x-Axis?

Consider the case where the middle of the circumvolve is on the x-centrality: (a, 0) is the center of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle. The equation of circle when the center is on the ten-axis is \((x - a)^2 + (y)^2 = r^2\)

How practise yous Graph a Circle Equation?

To graph a circle equation, first find out the coordinates of the center of the circumvolve and the radius of the circle with the help of the equation of the circle.

Then plot the center on a cartesian airplane and with the help of a compass mensurate the radius and depict the circle.

How do you Find the General Equation of Circumvolve?

If nosotros know the coordinates of the centre of a circle and the radius so we can discover the general equation of circle. For instance, the center of the circle is (1, 1) and the radius is 2 units and then the general equation of the circumvolve can be obtained past substituting the values of eye and radius.The general equation of the circle is \(x^2 + y^2 + Ax + By + C = 0\).

\(\text{A} = -ii \times one = -2\)
\(\text{B} = -two \times 1 = -2\)
\(\text{C} = 1^2 + 1^2 - two^2 = -ii\)

Hence the full general course of the equation of circle is \(10^ii + y^two - 2x - 2y - two = 0\).

How practise you lot Write the Standard Form of Equation of a Circumvolve?

The standard form of the equation of a circle is \((ten - x_1)^2 + (y - y_1)^ii = r^two\), where \((x_1, y_1)\) is the coordinate of the middle of the circle and \(r\) is the radius of the circumvolve

How practice you Go From Standard Form to a General Form of Equation of a Circle?

Let's catechumen the equation of circle: \({(x - ane)}^2 + {(y - two)}^2 = 4\) from standard form to gerenal class.

\({(x - 1)}^2 + {(y - 2)}^ii = 4 \\
x^2 + 1 - 2x + y^2 + 4 - 4y = 4 \\
x^2 + y^ii - 2x - 4y + 1 = 0 \)

The above form of the equation is the general class of the equation of circumvolve.

How exercise you Write the Standard Form of a Circumvolve Equation with Endpoints?

Let'southward accept the two endpoints of the diameter to be (i, 1), and (3, 3). Start, calculate the midpoint by using the section formula. The coordinates of the middle will be (2, 2). Secondly, calculate the radius by distance formula between (1, ane), and (two, 2). Radius is equal to \(\sqrt{2}\). Now, the equation of the circumvolve in standard class is \({(ten - 2)}^2 + {(y - two)}^2 = 2\).

What is the Polar Equation of a Circle?

The polar equation of the circle with the heart as the origin is, r = p, where p is the radius of the circle.

Source: https://www.cuemath.com/geometry/equation-of-circle/

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