Given a Point and a Radius R Draw a Circle
Equation of Circle
The equation of circle provides an algebraic way to describe a circle, given the center and the length of the radius of a circle. The equation of a circumvolve is dissimilar from the formulas that are used to summate the surface area or the circumference of a circle. This equation is used across many problems of circles in coordinate geometry.
To represent a circle on the Cartesian plane, nosotros crave the equation of the circle. A circumvolve tin be drawn on a slice of paper if we know its heart and the length of its radius. Similarly, on a Cartesian plane, nosotros tin draw a circle if we know the coordinates of the center and its radius. A circumvolve tin be represented in many forms:
- General grade
- Standard form
- Parametric form
- Polar form
In this article, let's learn about the equation of the circle, its various forms with graphs and solved examples.
| one. | What is the Equation of Circumvolve? |
| 2. | Different Forms of Equation of Circle |
| 3. | Equation of a Circle Formula |
| 4. | Derivation of Circle Equation |
| five. | Graphing the Equation of Circle |
| 6. | How to Detect Equation of Circumvolve? |
| 7. | Converting General Form to Standard Form |
| 8. | Converting Standard Form to General Form |
| 9. | FAQs on Equation of a Circumvolve |
What is the Equation of Circle?
An equation of a circle represents the position of a circle in a Cartesian plane. If we know the coordinates of the heart of the circle and the length of its radius, we tin write the equation of a circle. The equation of circle represents all the points that lie on the circumference of the circle.
A circle represents the locus of points whose distance from a fixed signal is a constant value. This fixed point is called the eye of the circumvolve and the abiding value is the radius r of the circle. The standard equation of a circle with center at \((x_1, y_1)\) and radius r is \( (x - x_1)^two + (y - y_1)^2 = r^2\).
Different Forms of Equation of Circle
An equation of circle represents the position of a circle on a cartesian aeroplane. A circumvolve can be fatigued on a slice of paper given its centre and the length of its radius. Using the equation of circle, one time nosotros find the coordinates of the eye of the circle and its radius, nosotros will be able to describe the circle on the cartesian plane. There are different forms to represent the equation of a circle,
- General class
- Standard form
- Parametric class
- Polar course
Let's await at the two common forms of the equation of circle-general form and standard class of the equation of circle hither along with the polar and parametric forms in particular.
General Equation of a Circumvolve
The general class of the equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0. This general class is used to discover the coordinates of the centre of the circle and the radius, where yard, f, c are constants. Unlike the standard course which is easier to understand, the general course of the equation of a circumvolve makes it difficult to find whatever meaningful backdrop about any given circle. So, nosotros volition be using the completing the square formula to make a quick conversion from the general form to the standard form.
Standard Equation of a Circle
The standard equation of a circumvolve gives precise information about the heart of the circle 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 circle.
The altitude between this betoken and the centre is equal to the radius of the circle. Let'southward apply the distance formula between these points.
\(\sqrt{(x - x_1)^2 + (y - y_1)^2} = r\)
Squaring both sides, we get the standard grade of the equation of the circle as:
\((ten - x_1)^2 + (y - y_1)^2 = r^2\)
Consider this example of an equation of circle (10 - 4)2 + (y - ii)2 = 36 is a circle centered at (4,ii) with a radius of vi.
Parametric Equation of a Circle
We know that the general form of the equation of a circle is xtwo + y2 + 2hx + 2ky + C = 0. We take a general bespeak on the purlieus of the circle, say (ten, y). The line joining this general point and the eye of the circle (-h, -1000) makes an angle of \(\theta\). The parametric equation of circle can be written as x2 + y2 + 2hx + 2ky + C = 0 where x = -h + rcosθ and y = -k + rsinθ.
Polar Equation of a Circle
The polar course of the equation of the circle is almost similar to the parametric class of the equation of circumvolve. Nosotros usually write the polar form of the equation of circumvolve for the circle centered at the origin. Permit's take a indicate P(rcosθ, rsinθ) on the purlieus of the circumvolve, 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 102 + ytwo = ptwo.
Substitute the value of x = rcosθ and y = rsinθ in the equation of circle.
(rcosθ)2 + (rsinθ)2 = p2
r2cos2θ + r2sin2θ = p2
rii(cos2θ + sin2θ) = pii
r2(1) = p2
r = p
where p is the radius of the circumvolve.
Instance: Find the equation of the circle in the polar form provided that the equation of the circle in standard grade is: x2 + y2 = 9.
Solution:
To find the equation of the circle in polar form, substitute the values of \(x\) and \(y\) with:
x = rcosθ
y = rsinθ
ten = rcosθ
y = rsinθ
10two + yii = ix
(rcosθ)ii + (rsinθ)2 = 9
r2cos2θ + riisintwoθ = 9
rtwo(cos2θ + sin2θ) = 9
rtwo(1) = ix
r = three
Equation of a Circumvolve Formula
The equation of a circle formula is used for calculating the equation of a circle. Nosotros tin can find the equation of whatever circle, given the coordinates of the center and the radius of the circle by applying the equation of circumvolve formula. The equation of circle formula is given as, \((10 - x_1)^two + (y - y_1)^2 = r^ii\).
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 (x, y) is an capricious point on the circumference of the circumvolve. The altitude between this signal and the heart is equal to the radius of the circumvolve. And then, let's apply the distance formula between these points.
\(\sqrt{(x - x_1)^2 + (y - y_1)^2} = r\).
Squaring both sides, nosotros go: \((x - x_1)^2 + (y - y_1)^2 = r^2\). So, the equation of a circle is given by:
\((x - x_1)^2 + (y - y_1)^2 = r^2\)
Instance: Using the equation of circle formula, find the center and radius of the circumvolve whose equation is (x - 1)2 + (y + 2)2 = ix.
Solution:
We will employ the circumvolve equation to determine the center and radius of the circle.
Comparing \((10 - ane)^two + (y + two)^ii = 9\) with \((10 - x_1)^ii + (y - y_1)^2 = r^2\), we get
\(x_1\) = 1, \(y_1\) = -two and r = 3
And then, the heart and radius are (1, -two) and 3 respectively.
Respond: The centre of the circle is (1, -two) and its radius is iii.
Graphing the Equation of Circle
In order to show how the equation of circle works, allow's graph the circle with the equation (ten -3)2 + (y - ii)2 = 9. Now, before graphing this equation, we need to make sure that the given equation matches the standard grade \((x - x_1)^2 + (y - y_1)^2 = r^2\).
- For this, we only need to alter the constant 9 to match with r2 as (x -three)ii + (y - 2)2 = 3ii.
- Here, we need to note that one of the common mistakes to commit is to consider \(x_{i}\) as -iii and \(y_{1}\) as -2.
- In the equation of circle, if the sign preceding \(x_{i}\) and \(y_{one}\) are negative, then \(x_{1}\) and \(y_{1}\) are positive values and vice versa.
- Here, \(x_{1}\) = iii, \(y_{ane}\) = 2 and r = 3
Thus, the circle represented by the equation (x -3)2 + (y - 2)two = 3two, has its center at (three, 2) and has a radius of 3. The beneath-given epitome shows the graph obtained from this equation of the circle.
How to Find Equation of Circle?
There are so many unlike 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 centre of a circle. There are certain special cases based on the position of the circle in the coordinate airplane. Let's acquire about the method to find the equation of circle for the full general and these special cases.
Equation of Circumvolve With Center at (x\(_1\), y\(_1\))
To write the equation of circle with eye at (x\(_1\), y\(_1\)), nosotros volition use the post-obit steps,
- Stride 1: Note down the coordinates of the eye of the circle(x\(_1\), y\(_1\)) and the radius 'r'.
- Step two: Use the equation of circumvolve formula, \(\sqrt{(x - x_1)^2 + (y - y_1)^ii} = r\).
- Footstep 3: Limited the respond in the required circumvolve equation form.
Equation of Circumvolve With Center at the Origin
The simplest example is where the circumvolve'south heart is at the origin (0, 0), whose radius is r. (10, y) is an arbitrary bespeak on the circumference of the circle.
The distance between this point and the middle is equal to the radius of the circle. Let's apply the distance formula between these points.
\( \sqrt{(ten - 0)^2 + (y - 0)^ii} = r\)
Squaring both sides, nosotros get:
\( (x - 0)^ii + (y - 0)^2 = r^ii\)
\( x^2 + y^two = r^2\)
Example: What will exist the equation of a circumvolve if its center is at the origin?
Solution:
The equation of a circle is given past \((ten - x_1)^2 + (y - y_1)^2 = r^two\).
If center is at origin, then \(x_1\)= 0 and \(y_1\)= 0.
Answer: The equation of the circle if its eye is at origin is 102+ ytwo= r2.
Equation of Circle With Middle on x-Axis
Consider the case where the center of the circle is on the x-axis: (a, 0) is the center of the circle with radius r. (x, y) is an arbitrary signal on the circumference of the circle.
The altitude betwixt this point and the center is equal to the radius of the circle. Let's use the distance formula between these points.
\(\sqrt{(x - a)^ii + (y - 0)^2} = r\)
Squaring both sides, we go:
\((x - a)^2 + (y - 0)^2 = r^2\)
\((x - a)^ii + (y)^two = r^2\)
Equation of Circle With Eye on Y-Axis
Consider the case where the center of the circle is on the y-centrality: (0, b) is the center of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle.
The distance betwixt this point and the middle is equal to the radius of the circle. Permit's utilise the distance formula between these points.
\( \sqrt{(x - 0)^2 + (y - b)^2} = r\)
Squaring both sides, we get:
\( (x - 0)^ii + (y - b)^ii = r^ii\)
\( (x)^two + (y - b)^2 = r^two\)
Equation of Circle Touching x-Axis
Consider the case where the circumference of the circle is touching the x-centrality at some betoken: (a, r) is the center of the circle with radius r. If a circumvolve touches the ten-axis, then the y-coordinate of the center of the circle is equal to the radius r.
(10, y) is an capricious signal on the circumference of the circle. The distance between this betoken and the centre is equal to the radius of the circle. Permit's use the distance formula between these points.
\( \sqrt{(ten - a)^ii + (y - r)^2} = r\)
Squaring both sides, we get:
\( (ten - a)^2 + (y - r)^2 = r^ii\)
Equation of Circle Touching y-Centrality
Consider the example where the circumference of the circle is touching the y-axis at some point: (r, b) is the center of the circle with radius r. If a circle touches the y-axis, then the x-coordinate of the center of the circumvolve is equal to the radius r.
(x, y) is an capricious point on the circumference of the circle. The altitude between this signal and the center is equal to the radius of the circle. Let's utilise the distance formula between these points.
\(\sqrt{(x - r)^2 + (y - b)^2} = r\)
Squaring both sides
\((ten - r)^two + (y - b)^two = r^2\)
Equation of Circumvolve Which Touches Both the Axes
Consider the instance where the circumference of the circle is touching both the axes at some bespeak: (r, r) is the heart of the circle with radius r. If a circle touches both the x-axis and y-axis, and so both the coordinates of the center of the circumvolve become equal to the radius (r, r).
(10, y) is an capricious indicate on the circumference of the circumvolve. The altitude between this bespeak and the center is equal to the radius of the circumvolve. Let's apply the altitude formula betwixt these points.
\(\sqrt{(x - r)^2 + (y - r)^2} = r\)
Squaring both sides
\((10 - r)^ii + (y - r)^2 = r^2\)
If a circumvolve touches both the axes, then consider the center of the circle to be (r,r), where r is the radius of the circle. Here, (r,r) can be positive as well every bit negative. For case, the radius of the circumvolve is 3 and it is touching both the axes, and so the coordinates of the center tin can be (3,iii), (3,−iii), (−3,3), or (−3,−iii).
Example: If the equation of circumvolve in general course is given equally \(x^2 + y^2 + 6x + 8y + 9 = 0\), find the coordinates of the center and the radius of the circumvolve.
Solution:
Given the equation of the circle \( x^2 + y^2 +6x + 8y + nine = 0\)
The full general form of the equation of the circle with heart \((x_1, y_1)\) and radius \(r\) is \( 10^2 + y^2 + Ax + By + C = 0\)
where \(A = -2x_1\)
\(B = -2y_1\)
\(C = {x_1}^2 + {y_1}^ii -r^2\)
From the equation of the circle \( x^two + y^two +6x + 8y + ix = 0\)
\(A = 6 \\
-2x_1 = vi \\
x_1 = -iii \\
B = viii \\
-2y_1 = 8 \\
y_1 = -4 \\
C = 9 \\
{x_1}^2 + {y_1}^2 -r^two = 9 \\
{-3}^two + {-4}^ii -r^2 = 9 \\
9 + 16 -r^ii = ix \\
r^ii = sixteen \\
r = four \)
Converting Full general Form to Standard Grade
This is the standard equation of circle, with radius r and center at (a,b): (x - a)ii + (y - b)2 = r2 and consider the general form every bit: x2 + y2 + 2gx + 2fy + c = 0. Hither are the steps to be followed to convert the general grade to the standard course:
Stride 1: Combine the similar terms and take the constant on the other side equally tentwo + 2gx + y2 + 2fy = - c -> (1)
Footstep 2: Employ the perfect foursquare identity (x + g)2 = xii + 2gx + 10002 to find the values of the expression xii + 2gx and y2 + 2fy equally:
(x + g)two = 102 + 2gx + kii ⇒ x2 + 2gx = (x + 1000)2 - g2 -> (ii)
(y + f)ii = y2 + 2fy + f2 ⇒ y2 + 2fy = (y + f)2 - f2 -> (three)
Substituting (two) and (3) in (one), nosotros get the equation as:
(ten+g)2 - thou2 + (y+f)ii - f2 = - c
(10+g)ii + (y+f)two = g2 + fii - c
Comparison this equation with the standard form: (x - a)2 + (y - b)ii = r2 nosotros get,
Center = (-m,-f) and radius = \(\sqrt{grand^2+f^2 - c}\)
Nosotros need to make sure that the coefficients of x2 and y2 are 1 before applying the formula.
Consider an case here to find the heart and radius of the circumvolve from the general equation of the circle: xii + y2 - 6x - 8y + 9 = 0.
The coordinates of the middle of the circumvolve can be plant every bit: (-k,-f). Hither g = -half-dozen/two = -3 and f = -8/two = -four. So, the center is (3,4).
Radius r = \(\sqrt{g^2+f^2 - c}\) = \(\sqrt{(-3)^{2}+(-4)^{2} - 9}\) = \(\sqrt{9 + 16 - 9}\) = \(\sqrt{16}\) = four. So, radius r = iv.
Converting Standard Course to General Course
We tin can use the algebraic identity formula of (a - b)2 = a2 + bii - 2ab to convert the standard course of equation of circle into the general class. Let's meet how to exercise this conversion. For this, expand the standard form of the equation of the circle every bit shown below, using the algebraic identities for squares:
\( (x - x_1)^2 + (y - y_1)^two = r^two\)
\( 10^2 +{x_1}^2 -2xx_1 + y^ii +{y_1}^2 -2yy_1 = r^ii\)
\( ten^2 + y^two - 2xx_1 - 2yy_1 + {x_1}^two + {y_1}^2 = r^2\)
\( x^2 + y^2 - 2xx_1 - 2yy_1 + {x_1}^2 + {y_1}^ii -r^2 = 0\)
Replace \(-2x_1\) with 2g, \(-2y_1\) with 2f, \( {x_1}^2 + {y_1}^2 -r^2\) with \(c\), nosotros get:
\( x^2 + y^2 + 2gx + 2fy + c = 0\)
Now, we get the general form of equation of circle as: \( x^2 + y^2 + 2gx + 2fy + c = 0\), where g, f, c are constants.
Related Articles on Equation of Circumvolve
Check out the post-obit pages related to the equation of circle
- Equation of a Circle Reckoner
- 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 circumvolve
- The general class of the equation of circle always has x2 + y2 in the beginning.
- If a circumvolve crosses both the axes, then there are four points of intersection of the circumvolve and the axes.
- If a circle touches both the axes, so there are merely 2 points of contact.
- If any equation is of the form \(x^2 + y^2 + axy + C = 0\), then it is non the equation of the circle. There is no \(xy\) term in the equation of circle.
- In polar form, the equation of circumvolve e'er represents in the class of \(r\) and \(\theta\).
- Radius is the altitude from the center to whatever 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 Circumvolve
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FAQs on Equation of Circle
What is the Equation of Circle in Geometry?
The equation of circle represents the locus of point whose distance from a fixed indicate is a constant value. This stock-still bespeak is called the eye of the circle and the constant value is the radius of the circumvolve. The standard equation of circle with centre at \((x_1, y_1)\) and radius r is \( (10 - x_1)^two + (y - y_1)^2 = r^2\).
What is the Equation of Circumvolve When the Heart Is at the Origin?
The simplest case is where the circle's centre is at the origin (0, 0), whose radius is r. (x, y) is an arbitrary point on the circumference of the circle. The equation of circumvolve when the center is at the origin is x2 + y2 = rii.
What is the Full general Equation of Circle?
The general form of the equation of circle is: tentwo + y2 + 2gx + 2fy + c = 0. This general grade of the equation of circle has a middle of (-m, -f), and the radius of the circle is r = \(\sqrt{grand^2 + f^2 - c}\).
What is the Parametric Equation of Circle?
The parametric equation of circle can be written as \(ten^ii + y^2 + 2hx + 2ky + C = 0\) where \(x = -h +rcos \theta\) and \(y = -k +rsin \theta\)
What is C in the General Equation of Circle?
The general form of the equation of circumvolve is: xii + y2 + 2gx + 2fy + c = 0. This general form is used to find the coordinates of the center of the circumvolve and the radius of the circle. Here, c is a abiding 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'southward look at the two common forms of the equation are:
- General Course tenii + y2 + 2gx + 2fy + C = 0
- Standard Class \((x - x_1)^2 + (y - y_1)^2 = r^2\)
What is the Equation of Circle When the Center is on 10-Axis?
Consider the case where the center of the circle is on the ten-axis: (a, 0) is the centre of the circle with radius r. (x, y) is an arbitrary bespeak on the circumference of the circle. The equation of circumvolve when the center is on the x-axis is \((x - a)^ii + (y)^2 = r^2\)
How practice y'all Graph a Circle Equation?
To graph a circle equation, first notice out the coordinates of the center of the circle and the radius of the circumvolve with the help of the equation of the circumvolve.
Then plot the centre on a cartesian plane and with the assistance of a compass measure the radius and draw the circle.
How do you Find the General Equation of Circle?
If we know the coordinates of the center of a circumvolve and the radius then nosotros tin find the general equation of circle. For example, the eye of the circumvolve is (1, 1) and the radius is 2 units so the full general equation of the circumvolve can be obtained by substituting the values of middle and radius.The general equation of the circle is \(x^ii + y^two + Ax + By + C = 0\).
\(\text{A} = -2 \times 1 = -two\)
\(\text{B} = -ii \times 1 = -two\)
\(\text{C} = ane^2 + one^2 - ii^ii = -2\)
Hence the full general course of the equation of circle is \(x^2 + y^2 - 2x - 2y - two = 0\).
How do yous Write the Standard Form of Equation of a Circle?
The standard grade of the equation of a circle is \((10 - x_1)^2 + (y - y_1)^2 = r^2\), where \((x_1, y_1)\) is the coordinate of the centre of the circle and \(r\) is the radius of the circle
How do you Get From Standard Course to a General Form of Equation of a Circle?
Let's convert the equation of circumvolve: \({(10 - 1)}^2 + {(y - 2)}^two = 4\) from standard form to gerenal course.
\({(x - ane)}^2 + {(y - 2)}^2 = 4 \\
x^ii + 1 - 2x + y^two + 4 - 4y = 4 \\
10^ii + y^two - 2x - 4y + one = 0 \)
The above grade of the equation is the general class of the equation of circle.
How do you Write the Standard Form of a Circle Equation with Endpoints?
Let's take the two endpoints of the bore to exist (ane, 1), and (iii, 3). First, summate the midpoint by using the section formula. The coordinates of the center will be (2, ii). Secondly, calculate the radius by altitude formula between (1, 1), and (2, 2). Radius is equal to \(\sqrt{2}\). At present, the equation of the circle in standard grade is \({(x - two)}^2 + {(y - ii)}^2 = two\).
What is the Polar Equation of a Circle?
The polar equation of the circle with the centre equally the origin is, r = p, where p is the radius of the circumvolve.
Source: https://www.cuemath.com/geometry/equation-of-circle/