Ecuación de un círculo

Contenido de página

• Ecuación de un círculo en forma estándar y Ejemplo 1
• Puntos sobre, dentro o fuera de un círculo y Ejemplo 2
• Recta tangente a un círculo y Ejemplo 3
• Forma general de la ecuación de un círculo y Ejemplo 4 , Ejemplo 5
• Tutoriales interactivos para explorar la ecuación de un círculo
• Referencias

Ecuación de un Círculo en Forma Estándar

A continuación se muestra un círculo con centro C dado por sus coordenadas C(h,k) . Por definición, todos los puntos M(x,y) del círculo están a la misma distancia del centro. En otras palabras, una circunferencia de centro C es el conjunto de todos los puntos que están a igual distancia del punto C . Esta distancia entre C y cualquier punto del círculo se llama radio y tiene una longitud r en el gráfico siguiente.


La distancia desde el centro C(h,k) hasta un punto M(x,y) en el círculo está dada por
Para encontrar la ecuación, usamos la definición para escribir que la distancia CM es igual al radio r
Trabajar con la raíz cuadrada agrega dificultades adicionales que se pueden evitar. La raíz cuadrada en la ecuación anterior se puede eliminar elevando al cuadrado ambos lados de la ecuación para obtener
Simplifique para obtener la ecuación estándar de un círculo con centro C(h,k) y radio r
\( \) \( \) \( \) \( \) \[ (x - h)^2 + (y - k)^2 = r^2 \]

Ejemplos

Ejemplo 1
Encuentra la ecuación estándar del círculo con centro C(2,5) y radio r = 2

Solución al Ejemplo 1
Dado el centro y el radio, la ecuación estándar del círculo viene dada por:
\( (x - 2)^2 + (y - 5)^2 = 2^2 \)

Puntos sobre, dentro o fuera de un círculo

Para saber si un punto dado está en un círculo, dentro o fuera de un círculo, comparamos el cuadrado de la distancia desde el centro del círculo hasta el punto dado con el cuadrado del radio. Usamos el cuadrado de la distancia en lugar de la distancia para evitar usar la raíz cuadrada.

For a circle of center \( C(h,k) \) and radius \( r \), point \( P \) with coordinates \( (x_0 , y_0) \)
1) is on the circle, if the following equality is satisfied:
\[ (x_0 - h)^2 + (y_0 - k)^2 = r^2 \]
2) is inside the circle, if the following inequality is satisfied:
\[ (x_0 - h)^2 + (y_0 - k)^2 \lt r^2 \]
3) is outside the circle, if the following inequality is satisfied:
\[ (x_0 - h)^2 + (y_0 - k)^2 \gt r^2 \]

Example 2
Equation of a circle and points inside, outside or on the circle
Which of the following points \( P_1(1,5) \) , \( P_2(2,3) \) and \( P_3(4,7) \) is inside, outside or on the circle given in example 1?

Solution to Example 2
The center of the circle in example 1 is at \( C(2,5) \) and has radius \( r = 2 \)
Find the square of the distance from the center of the circle to each of the given point and compare it to the square of the radius
square of distance from \( C \) to \( P_1 \) is given by: \( (2-1)^2 + (5-5)^2 = 1\) which is smaller than \( r^2 = 4\). Hence point \( P_1 \) is inside the circle.
square of distance from \( C \) to \( P_2 \) is given by: \( (2-2)^2 + (5-3)^2 = 4\) which is equal to \( r^2 = 4\). Hence point \( P_1 \) is on the circle.
square of distance from \( C \) to \( P_3 \) is given by: \( (2-4)^2 + (5-7)^2 = 8\) which is larger than \( r^2 = 4\). Hence point \( P_3 \) is outside the circle.
The circle and the three points are shown below and we can easily check the answer found above.

Tangent line to a circle

One of the important properties of a tangent line to a circle is that it is perependicular to the line through the center \( C \) and the point of tangency \( M \) as shown below.

Example 3 Equation of a circle and points inside, outside or on the circle
Find the equation of the tangent line to the circle with equation \( (x + 2)^2 + (y - 2)^2 = 5 \) at the point \( M(-4 , 3) \).
Solution to Example 3
Comparing the given equation to the general standard equation given above, we deduce that the center is at \( C(-2,2) \).
The slope \( m_1 \) of the line through \( C M \) is given by
\( m_1 = \dfrac{3-2}{-4-(-2)} = -\dfrac{1}{2} \)
Let \( m_2 \) be the slope of the tangent line. Since the tangent line and \( C M \) are perpendicular, the slopes \( m_1 \) and \( m_2 \) are related by
\( m_1 \times m_2 = -1 \)
which gives
\( (-\dfrac{1}{2}) \times m_2 = -1 \)
solve for \( m_2 \) to obtain
\( m_2 = 2 \)
we now know the slope \( m_2 \) and \( M(-4,3) \) the point of tangency through which the tangent line passes, we use the point slope formula to find the equation of the line tangent to the given circle
\( y - 3 = 2 (x - (-4)) \)
\( y = 2 x + 11 \)
As an exercise, plot the given circle and the equation of the tangent line found above and check graphically that they are tangent at the point (-4,3).

General Form of the Equation of a circle

Let us start with the standard equation of a circle
\( (x - h)^2 + (y - k)^2 = r^2 \)
Expand
\( x^2 - 2 h x + h^2 + y^2 - 2 k y + k^2 = r^2 \)
Let
\( A = - 2 h\), \( B = - 2 k \) and \( C = h^2 + k^2 - r^2\)
Substitute in the expanded equation to obtain the general form of the equatin of a circle.
\( x^2 + y^2 + A x + B y + C = 0 \)

Example 4 Find the equation of a circle given three points
Find the equation of the circle through the points \( P_1(6,4) \), \( P_2(-1,5) \) and \( P_3(2,-4) \).
Solution to Example 4
The coordinates of a point on a circle must satisfy the equation of the circle. We write that the coordinates of each of the given points satisfy the equation of the circle in the general form: \( x^2 + y^2 + A x + B y + C = 0 \).
Point \( P_1(6,4) \) is on the circle; substitute \( x \) by \( 6 \) and \( y \) by \( 4 \) in the equation, hence: \( 6^2 + 4^2 + 6 A + 4 B + C = 0 \)
Point \( P_2(-1,5) \) is on the circle; substitute \( x \) by \( -1 \) and \( y \) by \( 5 \) in the equation, hence: \( (-1)^2 + 5^2 - A + 5 B + C = 0 \)
Point \( P_3(2,-4) \) is on the circle; substitute \( x \) by \( 2 \) and \( y \) by \( -4 \) in the equation, hence: \( 2^2 + (-4)^2 + 2 A - 4 B + C = 0 \)

Solve the for (A,B,C) the system of equations obtained above and shown below in standard form
\( \begin{cases} 6 A + 4 B + C = -52\\ - A + 5 B + C = - 26 \\ 2 A - 4 B + C = -20 \end{cases} \)

Use any method to obtain the solution
\( A = -4\) , \( B = -2 \), \( C = -20 \)
We now substitute \( A \), \( B \) and \( C \) by their values and write the equation of the circle as follows:
\( x^2 + y^2 - 4 x - 2 y - 20 = 0 \)
The circle found above and the three points are shown in the graph below.

Example 5 Rewrite the general equation of a circle into standard form.
Rewrite the equation of the circle given by \( x^2 + y^2 + 6x - 2y + 5 = 0 \) and find its center and radius.
Solution to Example 5
Put, between parentheses, terms in \( x\) and \( x^2 \) together and the terms in \( y \) and \( y^2\) together
\( (x^2 + 6x) + (y^2 - 2y) + 5 = 0 \)
complete the square of each binomial within the parehtheses
\( (x^2 + 3)^2 - 3^2 + (y^2 - 1)^2 - (-1)^2 + 5 = 0 \)
Write in standard form
\( (x + 3)^2 + (y - 1)^2 = 5 \)
Compare the above equation to the standard one \( (x - h)^2 + (y - k)^2 = r^2\) and identify the coordinates \( h \) and \( k \) of the center of the circle and the radius \( r \).
\( h = - 3\) , \( k = 1\) and \( r^2 = 5 \)
Center of the circle has coordinates: \( (-3 , 1) \) and radius \( r = \sqrt 5 \)

Interactive Tutorials to Explore the Equation of a Circle

This is an HTML5 app to help explore the equation of a circle and the properties of the circle.

References and Links

Step by Step Maths Worksheets Solvers
Find x and y intercepts of Circles - Calculator
Tutorial on Equation of Circle
Maths Problems and Online Self Tests.