A circle is the locus of the points drawn at an equidistant from a fixed point. The fixed point is called its centre and the distance from the centre of the circle to the outer curve is its radius. The diameter is the line that divides the circle into two equal parts and is also equal to twice the radius.
Let’s understand what is the equation of a circle in coordinate geometry and how is it derived.
Equation of Circle in Coordinate Geometry
The equation of a circle provides an algebraic way to describe a circle, given the centre 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 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, we require the equation of the circle. A circle can be drawn on a piece of paper if we know its centre and the length of its radius. Similarly, on a Cartesian plane, we can draw a circle if we know the coordinates of the centre and its radius.
The equation of a 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 point is a constant value. This fixed point is called the centre of the circle and the constant value is the radius $r$ of the circle. The standard equation of a circle with centre at $(x_1, y_1)$ and radius $r$ is $(x – x_1)^2 + (y – y_1)^2 = r^2$.
Derivation of Equation of Circle in Coordinate Geometry
Let’s consider a circle with centre $\text{O}(x_1, y_1)$ and radius $r$.
Further consider a point on the circle $\text{P}(x, y)$. As this point moves along the circumference of the circle, it’s distance from the centre $\text{O}(x_1, y_1)$ remains fixed which is equal to the radius of the circle $r$.
The distance of the point $\text{P}(x, y)$ from the centre $\text{O}(x_1, y_1)$ can be calculated using the distance formula as $|\overline{\text{OP}}| = \sqrt{(x – x_1)^2 + (y – y_1)^2}$.
As the distance $|\overline{\text{OP}}|$ is equal to the radius of the circle $r$, therefore $\sqrt{(x – x_1)^2 + (y – y_1)^2} = r$.
Squaring both sides, we get
$(x – x_1)^2 + (y – y_1)^2 = r^2$ which is the equation of circle with centre $(x_1, y_1)$ and radius $r$.
When the centre of the circle is at the origin $(0, 0)$, the equation of circle becomes
$(x – 0)^2 + (y – 0)^2 = r^2$
$=> x^2 + y^2 = r^2$
Examples on Equation of Circle in Coordinate Geometry
Example 1: Find the equation of a circle with centre at $(5, 2)$ and radius $4$ units.
The equation of circle with centre at $(x_1, y_1)$ and radius $r$ is $(x – x_1)^2 + (y – y_1)^2 = r^2$.
Therefore the equation of a circle with centre at $(5, 2)$ and radius $4$ units is $(x – 5)^2 + (y – 2)^2 = 4^2$.
$=> x^2 – 10x + 25 + y^2 – 4y + 4 = 4$
$=> x^2 + y^2 – 10x – 4y + 25 + 4 – 4 = 0$
$=> x^2 + y^2 – 10x – 4y + 25 = 0$
$=> x^2 + y^2 – 10x – 4y = – 25$
Example 2: Find the equation of a circle with centre at $(-1, 3)$ and radius $2$ units.
The equation of circle with centre at $(x_1, y_1)$ and radius $r$ is $(x – x_1)^2 + (y – y_1)^2 = r^2$.
Therefore the equation of a circle with centre at $(-1, 3)$ and radius $2$ units is $(x – (-1))^2 + (y – 3)^2 = 2^2$.
$=> (x + 1)^2 + (y – 3)^2 = 4$
$=> x^2 + 2x + 1 + y^2 – 6y + 9 = 4$
$=> x^2 + y^2 + 2x – 6y + 10 = 4$
$=> x^2 + y^2 + 2x – 6y = 4 – 10$
$=> x^2 + y^2 + 2x – 6y = -6$
Equation of Circle With Coordinates of Diameter
Consider a circle with coordinates of the endpoints of the diameter as $(x_1, y_1)$ and $x_2, y_2$. Then the equation of circle is given by $(x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0$.
Derivation of Equation of Circle With Coordinates of Diameter
Consider the coordinates of endpoints of the diameter of a circle as $\text{A}(x_1, y_1)$ and $\text{B}(x_2, y_2)$. Since centre of a circle is the midpoint of its diameter, therefore coordinates of centre are $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Using distance formula, length of diameter = $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
Therefore radius of circle $r = \frac{1}{2} \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.
Thus equation of circle will be $\left(x – \frac{x_1 + x_2}{2} \right)^2 + \left(y – \frac{y_1 + y_2}{2} \right)^2 = \frac{1}{4} (x_2 – x_1)^2 + (y_2 – y_1)^2$
$\left(\frac{2x – x_1 – x_2}{2} \right)^2 + \left(\frac{2y – y_1 – y_2}{2} \right)^2 = \frac{1}{4} (x_2 – x_1)^2 + (y_2 – y_1)^2$
$=> \frac{(2x – x_1 – x_2)^2}{4} + \frac{(2y – y_1 – y_2)^2}{4} = \frac{1}{4} (x_2 – x_1)^2 + (y_2 – y_1)^2$
$=> (2x – x_1 – x_2)^2 + (2y – y_1 – y_2)^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2$
$=> 4x^2 + x_1^2 + x_2^2 – 4x x_1 + 2x_1 x_2 – 4x x_2 + 4y^2 + y_1^2 + y_2^2 – 4y y_1 + 2y_1 y_2 – 4y y_2$
$= x_2^2 – 2 x_1 x_2 + x_1^2 + y_2^2 – 2 y_1 y_2 + y_1^2$
$=> 4x^2 – 4x x_1 + 2x_1 x_2 – 4x x_2 + 4y^2 – 4y y_1 + 2y_1 y_2 – 4y y_2 = – 2 x_1 x_2 – 2 y_1 y_2$
$=> 4x^2 – 4x x_1 + 2x_1 x_2 – 4x x_2 + 4y^2 – 4y y_1 + 2y_1 y_2 – 4y y_2 + 2 x_1 x_2 + 2 y_1 y_2 = 0$
$=> 4x^2 – 4x x_1 + 4x_1 x_2 – 4x x_2 + 4y^2 – 4y y_1 + 4y_1 y_2 – 4y y_2 = 0$
$=> 4(x^2 – x x_1 + x_1 x_2 – x x_2 + y^2 – y y_1 + y_1 y_2 – y y_2) = 0$
$=> x^2 – x x_1 + x_1 x_2 – x x_2 + y^2 – y y_1 + y_1 y_2 – y y_2 = 0$
$=> (x^2 – x x_1) + (x_1 x_2 – x x_2) + (y^2 – y y_1) + (y_1 y_2 – y y_2) = 0$
$=> x(x – x_1) + x_2(x_1 – x) + y(y – y_1) + y_2(y_1 – y) = 0$
$=> x(x – x_1) – x_2(x – x_1) + y(y – y_1) – y_2(y – y_1) = 0$
$=> (x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0$
Examples on Equation of Circle With Coordinates of Diameter
Example 1: Find the equation of a circle whose coordinates of the endpoints of one of the diameters are $(-6, 4)$ and $(4, -6)$.
Equation of circle with ccordinates of endpoints of diameter $(x_1, y_1)$ and $(x_2, y_2)$ is $(x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0$.
Therefore the equation of circle with endpoints of the diameter $(-6, 4)$ and $(4, -6)$ is $(x – (-6))(x – 4) + (y – 4)(y – (-6)) = 0$.
$=> (x + 6)(x – 4) + (y – 4)(y + 6) = 0$
$=> x^2 – 4x + 6x – 24 + y^2 + 6y – 4y – 24 = 0$
$=> x^2 + 2x + y^2 + 2y – 48 = 0$
$=> x^2 + y^2 + 2x + 2y = 48$
Practice Problems
- Find the equation of a circle with centre and radius as
- Centre $(2, -4)$, Radius = $1$
- Centre $(5, 0)$, Radius = $4$
- Centre $\left(\frac{1}{2}, \frac{1}{4} \right)$, Radius = $3$
- Find the equation of a circle with endpoints of diameter as
- $(1, 1)$ and $(6, 6)$
- $(-5, 0)$ and $(5, 0)$
- $(0, 4)$ and $(0, -4)$
FAQs
What is the general equation of a circle?
The standard form for the equation of a circle is $(x − x_1)^2 + (y − y_1)^2 = r^2$, where $(x_1, y_1)$ is the centre and $r$ is the radius of a circle.
Is the equation of a circle a function?
A circle is not a function. The mathematical formula used to describe a circle is an equation, not one function. For a given set of inputs, a function must have at most one output.
What are the types of equations of a circle?
The two forms of the equation of a circle are
Standard Form: $(x − x_1)^2 + (y − y_1)^2 = r^2$
General Form: $x^2 + y^2 + 2gx + 2fy + c = 0$.
What is the equation of a circle when the centre is at the origin?
The simplest case is where the circle’s centre is at the origin $(0, 0)$, whose radius is $r$. The equation of a circle when the center is at the origin is $x^2 + y^2 = r^2$.
What is the equation of a circle when endpoints of a diameter are known?
The equation of a circle, when endpoints of a diameter are known, is $(x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of the diameter.
Conclusion
A circle is the locus of the points drawn at an equidistant from a fixed point also called a centre. The standard equation of a circle with centre at $(x_1, y_1)$ and radius $r$ is $(x – x_1)^2 + (y – y_1)^2 = r^2$. You can also find the equation of a circle when endpoints of diameter are known. The equation of a circle in this case is $(x – x_1)(x – x_2) + (y – y_1)(y – y_2) = 0$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of the diameter of a circle.
