In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. Show Danny B. Writing the equation of a circle centered at the origin given its radius or a point on the circle 1 Expert Answer
This calculator will find either the equation of the circle from the given parameters or the center, radius, diameter, circumference (perimeter), area, eccentricity, linear eccentricity, x-intercepts, y-intercepts, domain, and range of the entered circle. Also, it will graph the circle. Steps are available. Related calculators: Parabola Calculator, Ellipse Calculator, Hyperbola Calculator, Conic Section Calculator
Your InputFind the center, radius, diameter, circumference, area, eccentricity, linear eccentricity, x-intercepts, y-intercepts, domain, and range of the circle $$$x^{2} + y^{2} = 9$$$. SolutionThe standard form of the equation of a circle is $$$\left(x - h\right)^{2} + \left(y - k\right)^{2} = r^{2}$$$, where $$$\left(h, k\right)$$$ is the center of the circle and $$$r$$$ is the radius. Our circle in this form is $$$\left(x - 0\right)^{2} + \left(y - 0\right)^{2} = 3^{2}$$$. Thus, $$$h = 0$$$, $$$k = 0$$$, $$$r = 3$$$. The standard form is $$$x^{2} + y^{2} = 9$$$. The general form can be found by moving everything to the left side and expanding (if needed): $$$x^{2} + y^{2} - 9 = 0$$$. Center: $$$\left(0, 0\right)$$$. Radius: $$$r = 3$$$. Diameter: $$$d = 2 r = 6$$$. Circumference: $$$C = 2 \pi r = 6 \pi$$$. Area: $$$A = \pi r^{2} = 9 \pi$$$. Both eccentricity and linear eccentricity of a circle equal $$$0$$$. The x-intercepts can be found by setting $$$y = 0$$$ in the equation and solving for $$$x$$$ (for steps, see intercepts calculator). x-intercepts: $$$\left(-3, 0\right)$$$, $$$\left(3, 0\right)$$$ The y-intercepts can be found by setting $$$x = 0$$$ in the equation and solving for $$$y$$$: (for steps, see intercepts calculator). y-intercepts: $$$\left(0, -3\right)$$$, $$$\left(0, 3\right)$$$ The domain is $$$\left[h - r, h + r\right] = \left[-3, 3\right]$$$. The range is $$$\left[k - r, k + r\right] = \left[-3, 3\right]$$$. AnswerStandard form/equation: $$$x^{2} + y^{2} = 9$$$A. General form/equation: $$$x^{2} + y^{2} - 9 = 0$$$A. Graph: see the graphing calculator. Center: $$$\left(0, 0\right)$$$A. Radius: $$$3$$$A. Diameter: $$$6$$$A. Circumference: $$$6 \pi\approx 18.849555921538759$$$A. Area: $$$9 \pi\approx 28.274333882308139$$$A. Eccentricity: $$$0$$$A. Linear eccentricity: $$$0$$$A. x-intercepts: $$$\left(-3, 0\right)$$$, $$$\left(3, 0\right)$$$A. y-intercepts: $$$\left(0, -3\right)$$$, $$$\left(0, 3\right)$$$A. Domain: $$$\left[-3, 3\right]$$$A. Range: $$$\left[-3, 3\right]$$$A. No matter your proficiency in the geometry of a circle, the equation of the circle may still make your head spin. If this is the case, don't worry! This standard equation of a circle calculator will help you determine a circle's radius and center coordinates in a blink of an eye. If you are curious about how to find the equation of a circle, scroll down, and you'll find an explanation of the formula. If the geometric shape you're trying to analyze looks a bit squashed, it's probably an ellipse. In such a case, head straight to our ellipse calculator!
The standard equation of a circle is a way to describe all points lying on a circle with just one formula: (x−A)2+(y−B)2=r2\small (x - A)^2 + (y - B)^2 = r^2(x−A)2+(y−B)2=r2 (x,y)(x, y)(x,y) are the coordinates of any point lying on the circumference of the circle. It means that as long as you know all of the constants, you can input any xxx value to figure out the coordinates of any arbitrary point on the circle. (A,B)(A, B)(A,B) are the coordinates of the center point. Just be careful about the + and - signs here; for example, a circle with an equation of the form (x−3)2+(y+3)2=52(x - 3)^2 + (y + 3)^2 = 5^2(x−3)2+(y+3)2=52 will have a center point at (3,−3)(3, -3)(3,−3). r2r^2r2 is the radius of the circle raised to the power of two, so to find the radius, take the square root of this value. Our equation of the circle calculator finds not only these values but also the diameter, circumference, and area of the circle – all to save you time!
The equation of a circle may be defined in different ways; not only in the standard form shown above. A circle can also be described as the locus of all the points that satisfy the equations: x=rcos(α)y=rsin(α)\small \begin{align*} x &= r \cos(\alpha)\\[.5em] y &= r \sin(\alpha) \end{align*}xy=rcos(α)=rsin(α) where:
If you shift the center of the circle to (A, B) coordinates, you'll simply add them to the x and y coordinates to get the general parametric equation of a circle: x=A+rcos(α)y=B+rsin(α)\small \begin{align*} x &= A + r \cos(\alpha)\\[.5em] y &= B + r \sin(\alpha) \end{align*}xy=A+rcos(α)=B+rsin(α) As above, you should be careful about the signs of the center point.
To demonstrate that these two forms of the circle formula are equivalent, let's do a conversion between them.
x=rcos(t) ⟹ cos(t)=x/ry=rsin(t) ⟹ sin(t)=y/r\small\quad\ \ \begin{align*} x &= r \cos(t) \implies \cos(t) = x / r\\[.5em] y &= r \sin(t) \implies \sin(t) = y / r \end{align*} xy=rcos(t)⟹cos(t)=x/r=rsin(t)⟹sin(t)=y/r
sin2(t)+cos2(t)=1\small\quad\ \ \sin^2(t) + \cos^2(t) = 1 sin2(t)+cos2(t)=1
(y/r)2+(x/r)2=1\small\quad\ \ (y/r)^2 + (x/r)^2 = 1 (y/r)2+(x/r)2=1
y2/r2+x2/r2=1\small\quad\ \ y^2/r^2 + x^2/r^2 = 1 y2/r2+x2/r2=1
x2+y2=1\small\quad\ \ x^2 + y^2 = 1 x2+y2=1 Here we are! That's the standard equation of a circle.
OK, you've learned about different circle equations. But where are they come from? Let's find out!
Great, we have all the data needed. Now we only need to use one geometric law or a basic piece of trigonometry to tie the given values together:
(x2−x1)2+(y2−y1)2=r ⇓(x−A)2+(y−B)2=r\small\quad\ \ \begin{align*} &\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = r\\[1em] &\qquad\qquad\quad\ \ \Downarrow\\[1em] &\sqrt{(x - A)^2 + (y - B)^2} = r \end{align*} (x2−x1)2+(y2−y1)2=r ⇓(x−A)2+(y−B)2=r Square this obtained formula to get the standard equation of a circle: (x−A)2+(y−B)2=r2\small\quad\ \ (x - A)^2 + (y - B)^2 = r^2 (x−A)2+(y−B)2=r2 ∣OC∣2+∣CP∣2=r2\small\quad\ \ |OC|^2 + |CP|^2 = r^2 ∣OC∣2+∣CP∣2=r2 ∣OC∣=∣x−A∣∣CP∣=∣y−B∣\small\quad\ \ \begin{align*} |OC| &= |x - A|\\[.5em] |CP| &= |y - B| \end{align*} ∣OC∣∣CP∣=∣x−A∣=∣y−B∣ Coming back to the first equation, we get: ∣x−A∣2+∣y−B∣2=r2\small\quad\ \ |x - A|^2 + |y - B|^2 = r^2 ∣x−A∣2+∣y−B∣2=r2 This is because a2=(−a)2a^2 = (-a)^2a2=(−a)2. the above equation works for every quadrant, not only the first one. Then the equation can be simplified to a standard formula: (x−A)2+(y−B)2=r2\small\quad\ \ (x - A)^2 + (y - B)^2 = r^2 (x−A)2+(y−B)2=r2
cos(α)=OC/r ⟹ OC=rcos(α)sin(α)=CP/r ⟹ CP=rsin(α)\small\quad\ \ \begin{align*} \cos(\alpha) &= OC / r \implies \\[.5em]OC &= r\cos(\alpha)\\[1em] \sin(\alpha) &= CP / r \implies \\[.5em]CP &= r\sin(\alpha) \end{align*} cos(α)OCsin(α)CP=OC/r⟹=rcos(α)=CP/r⟹=rsin(α) Coordinates of point P may be expressed as: x=A+OC & y=B+CP⇓x=A+rcos(α)y=B+rsin(α)\small\quad\ \ \begin{align*} x = A + OC\ \ &\&\ \ y = B + CP\\[.5em] &\Downarrow\\[.5em] x = A &+ r\cos(\alpha)\\ y = B &+ r\sin(\alpha) \end{align*} x=A+OC x=Ay=B& y=B+CP⇓+rcos(α)+rsin(α) The two last equations represent the parametric equations of the circle. Now, as you know how to find the equation of a circle, give this calculator a try!
Quite often, you will come across another form of the circle equation. It looks like this: x2+y2+Dx+Ey+F=0\small x^2 + y^2 + Dx + Ey + F = 0x2+y2+Dx+Ey+F=0 This is the general form of a circle equation – the same as the standard form but expanded. It is possible to get to the standard form from here with just a few simple operations – all you have to do is follow the steps below!
x2+y2+4x−6y+8=0\small\quad\ \ x^2 + y^2 + 4x - 6y + 8 = 0 x2+y2+4x−6y+8=0
(x2+4x)+(y2−6y)+8=0\small\quad\ \ (x^2 + 4x) + (y^2 - 6y) + 8 = 0 (x2+4x)+(y2−6y)+8=0
(x2+4x)+(y2−6y)=−8\small\quad\ \ (x^2 + 4x) + (y^2 - 6y) = -8 (x2+4x)+(y2−6y)=−8
(x2+4x+4)+(y2−6y)=−8+4\small\quad\ \ \begin{align*} (x^2 + 4x + 4) + (y^2 - 6y) =\\ -8 + 4 \end{align*} (x2+4x+4)+(y2−6y)=−8+4
(x2+4x+4)+(y2−6y+9)=−8+4+9\small\quad\ \begin{align*} (x^2 + 4x + 4) + (y^2 - 6y + 9) =\\ -8 + 4 + 9 \end{align*} (x2+4x+4)+(y2−6y+9)=−8+4+9
(x+2)2+(y−3)2=5\small\quad\ (x + 2)^2 + (y - 3)^2 = 5 (x+2)2+(y−3)2=5 You're done – this is the standard form of a circle with a center at (−2,3)(-2, 3)(−2,3) and a radius equal to 5\sqrt55. Naturally, you don't have to go through this whole process by yourself. Use this equation of a circle calculator instead!
If you are still unsure how to use our tool, have a look at this simple example below.
In this calculator, we didn't put any units; you don't need them in many cases because you're working with the coordinates. If you need some units, e.g., inches, feet, or centimeters, just add them to the obtained results. Don't forget that the circles area will have squared units of length! Try our equation of a sphere calculator to do the same thing, but with spheres! |