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# Equation of the circle: Course, method and corrected exercises

This article aims to present how to calculate the equation of a circle and recognize which circle it is, through the course, examples and corrected exercises.

## Definition

The Cartesian equation of the circle in a plane is written in the form:

(x-x_A)^2 + (y-y_A)^2 = R^2

with:

• (xA,yA) the center of the circle
• R the radius of the circle

So if we know the radius of the circle and its center, it is easy to establish its Cartesian equation

## Corrected exercises and methods

### Find the equation of the circle from its center of its radius

We have the following statement: Consider the circle of radius 2 and radius (1,3). Find the equation of this circle. We have, according to the definition that the equation is written:

(x-1)^2 + (y-3) ^2 = 2^2

We will then develop this equation to simplify it:

x^2 -2x +1 +y^2 -6y +9 = 4

Then, we will simplify and put all the elements on the left:

x^2 +y^2 -2x-6y +6 = 0

We have therefore found the equation of the circle with center (1,3) and radius 2.

### Find the circle from its equation

Here is the typical statement:

Find the circle associated with the following equation:

x^2 - 2x + y^2 +6y = 0

For this, we will use the canonical form for the terms in x:

\begin{array}{ll} x^2 - 2 x & = x^2 - 2x +1 - 1 \\ &= (x-1)^2 -1 \end{array}

Then those in there:

\begin{array}{ll} y^2 +6 x & = y^2 +6y +9 - 9 \\ &= (y+3)^2 -9 \end{array}

If we combine the terms, we now have:

 (x-1)^2 -1 + (y+3)^2 -9 = 0

We pass the constants on the right:

 (x-1)^2 + (y+3)^2 = 10

Which we rewrite in the form:

 (x-1)^2 + (y+3)^2 = (\sqrt{10})^2

It is therefore the circle with center (1,-3) and radius √10

### Check that a point belongs to a circle

Statement: Let the circle of equation

(x-2)^2+(y-4)^2 = 25

Does the point (0;5) belong to the circle?

To do this, we replace x and y with the coordinates of our point. We then obtain:

(0-2)^2 +(5-4)^2 = 5 \neq 25

So the point does not belong to the circle because the left member is not equal to 25.

## Exercices

### Exercise 1

Write the equation of the circle from the coordinates of the center Ω​ and the radius R:

\begin{array}{l} 1)\ \Omega = (1;6) , R= 4\\ 2)\ \Omega = (5;3) , R= 2\\ 3)\ \Omega = (7 ;4) , R= 6\\ \end{array}

### Exercise 2

From the diameter [AB], give the equation of the circle:

\begin{array}{l} 1)\ A= (1;6) , B=(2;3)\\ 2)\ A= (2;5) , B=(3;2)\\ 3) \ A= (4;5) , B=(1;1)\\ \end{array}

### Exercise 3

Write the equation of the circle from the coordinates of the center Ω​ and a point A:

\begin{array}{l} 1)\ \Omega = (1;6) , A= (1;3)\\ 2)\ \Omega = (2;4) , A= (3;1)\\ 3)\ \Omega = (4;2) , A= (5;5)\\ \end{array}

### Exercise 4

In an orthonormal reference of the plane, determine the coordinates of the center and the radius of the circle whose equation is:

x^2+y^2 -12x + 4y -3 = 0

### Exercise 5

In an orthonormal reference of the plane, determine the coordinates of the center and the radius of the circle whose equation is:

\begin{array}{l} 1)\ x^2+y^2 = 0\\ 2)\ x^2 + y^2 = - 1\\ 3)\ (x-3)(x+2) +(y-5)(y+6)=0 \end{array}

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