Equations of circles (centre at origin and at (h, k))
Equations of Circles
Equations of Circles is a topic in Circles in the Common Core State Standards. It is aligned to Standard G-GPE.1, which requires students to derive the equation of a circle given its center and radius using the Pythagorean theorem.
The equation of a circle with center \((h,k)\) and radius \(r\) is \((x-h)^2+(y-k)^2=r^2\).
Theory
A circle with center \((h,k)\) and radius \(r\) has the standard-form equation
Centered at the origin, it simplifies to \(x^2+y^2=r^2\). Given a general-form equation, complete the square in \(x\) and \(y\) to find the center and radius.
Standard form and the origin case:
How to work with circle equations
- Write: substitute \(h,k,r\) into the standard form.
- Read: the center is \((h,k)\) with opposite signs; radius \(=\sqrt{\text{right side}}\).
- General form: complete the square in \(x\) and \(y\).
- Balance the equation when adding to complete the square.
Substitute into \((x-h)^2+(y-k)^2=r^2\).
| \((x-2)^2+(y+3)^2\) | \(=\) | \(25\) |
Read \(h,k\) (with opposite signs) and \(r=\sqrt{36}\).
| \(\text{center}\) | \(=\) | \((-1,4)\) |
| \(r\) | \(=\) | \(\sqrt{36}=6\) |
Group and complete the square in \(x\) and \(y\).
| \((x^2-6x)+(y^2+4y)\) | \(=\) | \(12\) |
| \((x-3)^2+(y+2)^2\) | \(=\) | \(12+9+4=25\) |
Center \((3,-2)\), radius \(5\).
With center \((0,0)\), the equation is \(x^2+y^2=r^2\).
| \(x^2+y^2\) | \(=\) | \(49\) |
Common pitfalls
Frequently asked questions
What is the standard equation of a circle?
\((x-h)^2+(y-k)^2=r^2\), with center \((h,k)\) and radius \(r\).
How do you find the center and radius from the equation?
The center is \((h,k)\) with signs opposite those in the equation; the radius is the square root of the right side.
How do you find a circle from its general form?
Group the \(x\) and \(y\) terms and complete the square in each to reach standard form.
What is the equation of a circle centered at the origin?
\(x^2+y^2=r^2\).