General Equation of a Circle Drawer
Equation of Circumvolve
The equation of circumvolve provides an algebraic way to describe a circumvolve, given the heart and the length of the radius of a circle. The equation of a circle is different from the formulas that are used to calculate the expanse or the circumference of a circumvolve. This equation is used beyond many problems of circles in coordinate geometry.
To represent a circle on the Cartesian plane, we crave the equation of the circle. A circle can be fatigued on a slice of paper if nosotros know its middle and the length of its radius. Similarly, on a Cartesian plane, we can draw a circle if we know the coordinates of the center and its radius. A circle tin be represented in many forms:
- General class
- Standard grade
- Parametric form
- Polar class
In this article, let's learn about the equation of the circle, its various forms with graphs and solved examples.
| 1. | What is the Equation of Circle? |
| ii. | Different Forms of Equation of Circumvolve |
| 3. | Equation of a Circle Formula |
| 4. | Derivation of Circle Equation |
| v. | Graphing the Equation of Circle |
| vi. | How to Find Equation of Circumvolve? |
| vii. | Converting General Form to Standard Form |
| 8. | Converting Standard Class to Full general Course |
| 9. | FAQs on Equation of a Circle |
What is the Equation of Circumvolve?
An equation of a circle represents the position of a circle in a Cartesian airplane. If we know the coordinates of the middle of the circle and the length of its radius, we can write the equation of a circle. The equation of circle represents all the points that lie on the circumference of the circumvolve.
A circumvolve represents the locus of points whose altitude from a fixed signal is a constant value. This fixed point is called the center of the circle and the constant value is the radius r of the circle. The standard equation of a circumvolve with center at \((x_1, y_1)\) and radius r is \( (x - x_1)^2 + (y - y_1)^two = r^2\).
Different Forms of Equation of Circumvolve
An equation of circle represents the position of a circle on a cartesian plane. A circumvolve tin can be drawn on a piece of paper given its center and the length of its radius. Using the equation of circle, once we find the coordinates of the center of the circle and its radius, we will exist able to draw the circle on the cartesian aeroplane. There are dissimilar forms to represent the equation of a circle,
- General form
- Standard course
- Parametric form
- Polar form
Let'south expect at the two common forms of the equation of circumvolve-general form and standard class of the equation of circle here along with the polar and parametric forms in detail.
General Equation of a Circle
The general form of the equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0. This full general class is used to discover the coordinates of the centre of the circumvolve and the radius, where 1000, f, c are constants. Dissimilar the standard form which is easier to understand, the general form of the equation of a circle makes it difficult to find whatsoever meaningful properties virtually whatever given circle. So, we will exist using the completing the foursquare formula to brand a quick conversion from the general grade to the standard form.
Standard Equation of a Circle
The standard equation of a circumvolve gives precise information almost the center of the circumvolve and its radius and therefore, it is much easier to read the heart 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 \( (ten - x_1)^two + (y - y_1)^ii = r^ii\), where (x, y) is an arbitrary point on the circumference of the circle.
The distance betwixt this betoken and the centre is equal to the radius of the circle. Let's use the distance formula between these points.
\(\sqrt{(x - x_1)^two + (y - y_1)^2} = r\)
Squaring both sides, nosotros become the standard form of the equation of the circle as:
\((ten - x_1)^two + (y - y_1)^2 = r^2\)
Consider this case of an equation of circumvolve (x - four)2 + (y - 2)2 = 36 is a circle centered at (4,2) with a radius of vi.
Parametric Equation of a Circle
We know that the full general form of the equation of a circumvolve is x2 + y2 + 2hx + 2ky + C = 0. Nosotros accept a general indicate on the purlieus of the circle, say (ten, y). The line joining this full general betoken and the heart of the circle (-h, -k) makes an angle of \(\theta\). The parametric equation of circumvolve tin be written as 102 + ytwo + 2hx + 2ky + C = 0 where x = -h + rcosθ and y = -k + rsinθ.
Polar Equation of a Circumvolve
The polar form of the equation of the circle is almost similar to the parametric form of the equation of circle. We unremarkably write the polar form of the equation of circle for the circle centered at the origin. Let'due south take a point P(rcosθ, rsinθ) on the boundary of the circle, where r is the distance of the point from the origin. We know that the equation of circle centered at the origin and having radius 'p' is x2 + y2 = p2.
Substitute the value of x = rcosθ and y = rsinθ in the equation of circle.
(rcosθ)2 + (rsinθ)2 = p2
riicos2θ + r2sintwoθ = ptwo
rtwo(cos2θ + sin2θ) = pii
rii(1) = p2
r = p
where p is the radius of the circle.
Example: Detect the equation of the circle in the polar form provided that the equation of the circle in standard form is: 10two + ytwo = 9.
Solution:
To find the equation of the circle in polar form, substitute the values of \(x\) and \(y\) with:
x = rcosθ
y = rsinθ
x = rcosθ
y = rsinθ
x2 + ytwo = 9
(rcosθ)two + (rsinθ)2 = ix
riicos2θ + r2siniiθ = nine
r2(cos2θ + sin2θ) = 9
r2(1) = nine
r = 3
Equation of a Circle Formula
The equation of a circle formula is used for calculating the equation of a circumvolve. We can notice the equation of any circle, given the coordinates of the center and the radius of the circle by applying the equation of circle formula. The equation of circumvolve formula is given as, \((10 - x_1)^ii + (y - y_1)^two = r^2\).
where,\((x_1, y_1)\) is the center of the circle with radius r and (10, y) is an arbitrary signal on the circumference of the circle.
Derivation of Circle Equation
Given that \((x_1, y_1)\) is the center of the circumvolve with radius r and (x, y) is an arbitrary point on the circumference of the circle. The distance betwixt this point and the heart is equal to the radius of the circle. And so, permit's employ the distance formula betwixt these points.
\(\sqrt{(x - x_1)^2 + (y - y_1)^ii} = r\).
Squaring both sides, we go: \((x - x_1)^two + (y - y_1)^2 = r^2\). So, the equation of a circle is given by:
\((x - x_1)^2 + (y - y_1)^2 = r^two\)
Instance: Using the equation of circle formula, find the center and radius of the circle whose equation is (10 - i)2 + (y + 2)2 = ix.
Solution:
We will use the circle equation to determine the center and radius of the circle.
Comparing \((10 - one)^2 + (y + 2)^ii = 9\) with \((x - x_1)^2 + (y - y_1)^2 = r^ii\), we get
\(x_1\) = 1, \(y_1\) = -2 and r = iii
And so, the center and radius are (one, -ii) and 3 respectively.
Answer: 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 circle works, allow's graph the circle with the equation (10 -iii)two + (y - ii)2 = 9. Now, before graphing this equation, we need to make sure that the given equation matches the standard class \((ten - x_1)^2 + (y - y_1)^2 = r^2\).
- For this, we simply demand to change the constant 9 to match with r2 every bit (x -3)2 + (y - 2)two = threetwo.
- Here, we need to notation that one of the common mistakes to commit is to consider \(x_{one}\) equally -3 and \(y_{1}\) as -two.
- In the equation of circle, if the sign preceding \(x_{1}\) and \(y_{1}\) are negative, then \(x_{1}\) and \(y_{one}\) are positive values and vice versa.
- Hither, \(x_{1}\) = three, \(y_{1}\) = 2 and r = 3
Thus, the circle represented by the equation (10 -3)ii + (y - 2)2 = 32, has its center at (iii, 2) and has a radius of 3. The below-given image shows the graph obtained from this equation of the circumvolve.
How to Find Equation of Circle?
In that location are so many different ways of representing the equation of circle depending on the position of the circle on the cartesian airplane. We have studied the forms to represent the equation of circle for given coordinates of center of a circle. There are certain special cases based on the position of the circle in the coordinate plane. Let'due south larn about the method to detect the equation of circumvolve for the full general and these special cases.
Equation of Circle With Eye at (x\(_1\), y\(_1\))
To write the equation of circle with center at (x\(_1\), y\(_1\)), we will use the post-obit steps,
- Footstep ane: Annotation downward the coordinates of the centre of the circle(x\(_1\), y\(_1\)) and the radius 'r'.
- Pace two: Apply the equation of circle formula, \(\sqrt{(x - x_1)^two + (y - y_1)^two} = r\).
- Step 3: Express the reply in the required circle equation course.
Equation of Circle With Center at the Origin
The simplest case is where the circle's center is at the origin (0, 0), whose radius is r. (ten, y) is an arbitrary point on the circumference of the circle.
The distance between this point and the center is equal to the radius of the circle. Let'south apply the distance formula betwixt these points.
\( \sqrt{(x - 0)^2 + (y - 0)^2} = r\)
Squaring both sides, we get:
\( (x - 0)^2 + (y - 0)^2 = r^2\)
\( x^2 + y^2 = r^ii\)
Instance: What volition be the equation of a circumvolve if its center is at the origin?
Solution:
The equation of a circle is given by \((x - x_1)^ii + (y - y_1)^2 = r^two\).
If center is at origin, so \(x_1\)= 0 and \(y_1\)= 0.
Answer: The equation of the circumvolve if its center is at origin is xii+ ytwo= r2.
Equation of Circumvolve With Centre on x-Centrality
Consider the instance where the center of the circumvolve is on the x-axis: (a, 0) is the eye of the circumvolve with radius r. (x, y) is an capricious bespeak on the circumference of the circumvolve.
The distance betwixt this point and the center is equal to the radius of the circumvolve. Let's apply the distance formula between these points.
\(\sqrt{(x - a)^2 + (y - 0)^2} = r\)
Squaring both sides, nosotros get:
\((x - a)^ii + (y - 0)^2 = r^two\)
\((x - a)^ii + (y)^2 = r^ii\)
Equation of Circumvolve With Centre on Y-Axis
Consider the case where the center of the circumvolve is on the y-axis: (0, b) is the centre of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle.
The distance betwixt this bespeak and the heart is equal to the radius of the circle. Let's apply the distance formula betwixt these points.
\( \sqrt{(x - 0)^2 + (y - b)^2} = r\)
Squaring both sides, we get:
\( (ten - 0)^2 + (y - b)^2 = r^2\)
\( (ten)^2 + (y - b)^2 = r^2\)
Equation of Circle Touching x-Centrality
Consider the case where the circumference of the circle is touching the ten-axis at some indicate: (a, r) is the eye of the circle with radius r. If a circle touches the 10-centrality, then the y-coordinate of the center of the circumvolve is equal to the radius r.
(ten, y) is an arbitrary point on the circumference of the circle. The distance between this indicate and the center is equal to the radius of the circle. Let's utilise the distance formula betwixt these points.
\( \sqrt{(x - a)^2 + (y - r)^2} = r\)
Squaring both sides, we become:
\( (x - a)^2 + (y - r)^two = r^2\)
Equation of Circle Touching y-Axis
Consider the case where the circumference of the circle is touching the y-axis at some point: (r, b) is the middle of the circle with radius r. If a circle touches the y-centrality, and then the x-coordinate of the center of the circle is equal to the radius r.
(x, y) is an arbitrary indicate on the circumference of the circle. The distance between this betoken and the center is equal to the radius of the circle. Let'south use the distance formula between these points.
\(\sqrt{(x - r)^2 + (y - b)^2} = r\)
Squaring both sides
\((x - r)^2 + (y - b)^2 = r^2\)
Equation of Circle Which Touches Both the Axes
Consider the case where the circumference of the circle is touching both the axes at some point: (r, r) is the heart of the circle with radius r. If a circumvolve touches both the ten-axis and y-centrality, then both the coordinates of the center of the circumvolve become equal to the radius (r, r).
(x, y) is an arbitrary point on the circumference of the circle. The distance between this point and the eye is equal to the radius of the circumvolve. Permit's use the distance formula between these points.
\(\sqrt{(ten - r)^two + (y - r)^2} = r\)
Squaring both sides
\((x - r)^2 + (y - r)^2 = r^2\)
If a circle touches both the axes, and so consider the center of the circumvolve to be (r,r), where r is the radius of the circle. Here, (r,r) can be positive as well every bit negative. For instance, the radius of the circle is 3 and it is touching both the axes, so the coordinates of the center tin exist (3,3), (iii,−3), (−3,three), or (−3,−3).
Example: If the equation of circumvolve in full general form is given equally \(ten^ii + y^two + 6x + 8y + 9 = 0\), find the coordinates of the center and the radius of the circle.
Solution:
Given the equation of the circumvolve \( x^2 + y^2 +6x + 8y + nine = 0\)
The general form of the equation of the circle with center \((x_1, y_1)\) and radius \(r\) is \( x^2 + y^2 + Ax + Past + C = 0\)
where \(A = -2x_1\)
\(B = -2y_1\)
\(C = {x_1}^ii + {y_1}^2 -r^2\)
From the equation of the circle \( x^2 + y^ii +6x + 8y + ix = 0\)
\(A = 6 \\
-2x_1 = 6 \\
x_1 = -3 \\
B = 8 \\
-2y_1 = 8 \\
y_1 = -four \\
C = 9 \\
{x_1}^2 + {y_1}^2 -r^two = 9 \\
{-3}^2 + {-4}^2 -r^ii = 9 \\
ix + 16 -r^ii = 9 \\
r^2 = sixteen \\
r = 4 \)
Converting General Form to Standard Form
This is the standard equation of circle, with radius r and heart at (a,b): (x - a)ii + (y - b)2 = r2 and consider the general grade every bit: x2 + y2 + 2gx + 2fy + c = 0. Here are the steps to exist followed to convert the full general form to the standard course:
Step ane: Combine the like terms and take the constant on the other side as xii + 2gx + yii + 2fy = - c -> (ane)
Footstep 2: Use the perfect square identity (x + grand)two = x2 + 2gx + thousand2 to find the values of the expression x2 + 2gx and y2 + 2fy as:
(10 + g)ii = x2 + 2gx + gtwo ⇒ ten2 + 2gx = (x + yard)2 - g2 -> (2)
(y + f)2 = y2 + 2fy + f2 ⇒ y2 + 2fy = (y + f)2 - f2 -> (three)
Substituting (two) and (three) in (i), we become the equation equally:
(x+g)ii - one thousandii + (y+f)2 - f2 = - c
(x+one thousand)2 + (y+f)2 = g2 + f2 - c
Comparing this equation with the standard class: (x - a)2 + (y - b)2 = r2 we go,
Heart = (-g,-f) and radius = \(\sqrt{1000^2+f^2 - c}\)
We need to brand sure that the coefficients of x2 and y2 are ane earlier applying the formula.
Consider an example here to find the center and radius of the circle from the full general equation of the circle: x2 + y2 - 6x - 8y + ix = 0.
The coordinates of the middle of the circle tin can be plant equally: (-g,-f). Here thousand = -6/2 = -iii and f = -8/ii = -4. So, the center is (3,4).
Radius r = \(\sqrt{g^2+f^two - c}\) = \(\sqrt{(-3)^{two}+(-iv)^{2} - 9}\) = \(\sqrt{9 + 16 - ix}\) = \(\sqrt{16}\) = 4. So, radius r = 4.
Converting Standard Form to General Course
We tin can employ the algebraic identity formula of (a - b)2 = a2 + b2 - 2ab to convert the standard class of equation of circumvolve into the full general grade. Allow'south see how to do this conversion. For this, expand the standard form of the equation of the circle every bit shown below, using the algebraic identities for squares:
\( (ten - x_1)^2 + (y - y_1)^2 = r^2\)
\( 10^2 +{x_1}^2 -2xx_1 + y^ii +{y_1}^two -2yy_1 = r^2\)
\( x^two + y^two - 2xx_1 - 2yy_1 + {x_1}^two + {y_1}^2 = r^2\)
\( x^2 + y^2 - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^2 -r^2 = 0\)
Replace \(-2x_1\) with 2g, \(-2y_1\) with 2f, \( {x_1}^2 + {y_1}^ii -r^2\) with \(c\), we go:
\( x^two + y^two + 2gx + 2fy + c = 0\)
Now, we go the general grade of equation of circle as: \( x^2 + y^2 + 2gx + 2fy + c = 0\), where g, f, c are constants.
Related Articles on Equation of Circle
Cheque out the post-obit pages related to the equation of circle
- Equation of a Circle Calculator
- Circumference of a Circle
- All Circle Formulas
- Ratio of Circumference to Diameter
Important Notes on Equation of Circle
Here is a list of a few points that should be remembered while studying the equation of circle
- The general form of the equation of circle always has x2 + y2 in the beginning.
- If a circumvolve crosses both the axes, then in that location are 4 points of intersection of the circle and the axes.
- If a circle touches both the axes, and then at that place are just ii points of contact.
- If any equation is of the form \(ten^two + y^ii + axy + C = 0\), 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 circumvolve always represents in the form of \(r\) and \(\theta\).
- Radius is the distance from the middle to whatsoever betoken on the boundary of the circle. Hence, the value of the radius of the circle is always positive.
Examples on Circle Equations
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Practice Questions on Equation of Circle
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FAQs on Equation of Circle
What is the Equation of Circle in Geometry?
The equation of circumvolve represents the locus of point whose distance from a fixed point is a constant value. This fixed point is called the center of the circle and the constant value is the radius of the circle. The standard equation of circle with heart at \((x_1, y_1)\) and radius r is \( (10 - x_1)^2 + (y - y_1)^2 = r^2\).
What is the Equation of Circle When the Center Is at the Origin?
The simplest case is where the circle's middle is at the origin (0, 0), whose radius is r. (x, y) is an capricious betoken on the circumference of the circle. The equation of circle when the middle is at the origin is ten2 + y2 = r2.
What is the General Equation of Circle?
The general form of the equation of circle is: 102 + ytwo + 2gx + 2fy + c = 0. This general form of the equation of circumvolve has a center of (-k, -f), and the radius of the circle is r = \(\sqrt{g^2 + f^2 - c}\).
What is the Parametric Equation of Circle?
The parametric equation of circle tin can be written as \(10^2 + y^2 + 2hx + 2ky + C = 0\) where \(10 = -h +rcos \theta\) and \(y = -yard +rsin \theta\)
What is C in the Full general Equation of Circle?
The general form of the equation of circumvolve is: xtwo + y2 + 2gx + 2fy + c = 0. This general form is used to find the coordinates of the middle of the circle and the radius of the circle. Hither, c is a constant term, and the equation having c value represents a circle that is non passing through the origin.
What are the Various Forms of Equations of a Circle?
Let's look at the two common forms of the equation are:
- Full general Form x2 + y2 + 2gx + 2fy + C = 0
- Standard Form \((x - x_1)^2 + (y - y_1)^2 = r^2\)
What is the Equation of Circle When the Heart is on x-Axis?
Consider the instance where the heart of the circumvolve is on the x-axis: (a, 0) is the centre of the circumvolve with radius r. (x, y) is an capricious point on the circumference of the circumvolve. The equation of circumvolve when the middle is on the x-centrality is \((10 - a)^2 + (y)^2 = r^2\)
How practise you 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 circumvolve.
Then plot the middle on a cartesian plane and with the help of a compass measure the radius and draw the circle.
How practise you Find the General Equation of Circumvolve?
If we know the coordinates of the center of a circle and the radius and then we can find the full general equation of circle. For example, the center of the circumvolve is (one, one) and the radius is two units and then the general equation of the circle can be obtained by substituting the values of center and radius.The general equation of the circle is \(ten^2 + y^2 + Ax + By + C = 0\).
\(\text{A} = -2 \times ane = -2\)
\(\text{B} = -2 \times ane = -2\)
\(\text{C} = one^ii + 1^ii - 2^ii = -2\)
Hence the full general form of the equation of circle is \(x^two + y^two - 2x - 2y - ii = 0\).
How do you Write the Standard Form of Equation of a Circle?
The standard form of the equation of a circle is \((x - x_1)^2 + (y - y_1)^two = r^two\), where \((x_1, y_1)\) is the coordinate of the centre of the circle and \(r\) is the radius of the circumvolve
How exercise you lot Become From Standard Class to a General Course of Equation of a Circle?
Let'south catechumen the equation of circle: \({(10 - one)}^2 + {(y - 2)}^2 = iv\) from standard form to gerenal form.
\({(10 - i)}^two + {(y - 2)}^2 = 4 \\
x^2 + 1 - 2x + y^ii + 4 - 4y = 4 \\
10^2 + y^2 - 2x - 4y + one = 0 \)
The above form of the equation is the general grade of the equation of circumvolve.
How do you Write the Standard Grade of a Circumvolve Equation with Endpoints?
Allow's take the two endpoints of the diameter to be (1, 1), and (iii, three). First, calculate the midpoint past using the section formula. The coordinates of the center will exist (2, 2). Secondly, calculate the radius by distance formula between (ane, 1), and (ii, ii). Radius is equal to \(\sqrt{ii}\). Now, the equation of the circle in standard form is \({(10 - 2)}^2 + {(y - 2)}^ii = two\).
What is the Polar Equation of a Circumvolve?
The polar equation of the circle with the middle as the origin is, r = p, where p is the radius of the circle.
Source: https://www.cuemath.com/geometry/equation-of-circle/