2.3 Circles

Equation of a Circle: MPM2D Grade 10 Mathematics

Equation of a Circle: MPM2D Grade 10 Mathematics

1. Introduction to the Equation of a Circle

A circle is a set of all points in a plane that are at a fixed distance (called the radius) from a fixed point (called the center). The standard form of the equation of a circle depends on the location of its center.

1.1. Equation of a Circle with Center at the Origin

If the center of the circle is at the origin (0,0), the equation of the circle is:

\[ x^2 + y^2 = r^2 \]

Example:

Consider a circle with a radius of 5 units centered at the origin.

The equation is:

\[ x^2 + y^2 = 5^2 \]

\[ x^2 + y^2 = 25 \]

1.2. Equation of a Circle with Center at Any Other Point

If the center of the circle is at a point \((h, k)\), the equation of the circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Example:

Consider a circle with a radius of 4 units and center at \((3, -2)\).

The equation is:

\[ (x - 3)^2 + (y + 2)^2 = 4^2 \]

\[ (x - 3)^2 + (y + 2)^2 = 16 \]

2. Finding the Radius of a Circle

2.1. Finding the Radius from the Equation

Given the equation of a circle, the radius can be found by isolating \( r \).

Example:

Given the equation \( (x - 1)^2 + (y + 4)^2 = 9 \),

the radius is:

\[ r = \sqrt{9} \]

\[ r = 3 \]

2.2. Finding the Radius from Given Coordinates

Given the center \((h, k)\) and a point \((x_1, y_1)\) on the circle, the radius can be found using the distance formula:

\[ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} \]

Example:

For a circle with center \((2, 3)\) and passing through the point \((5, 7)\):

\[ r = \sqrt{(5 - 2)^2 + (7 - 3)^2} \]

\[ r = \sqrt{3^2 + 4^2} \]

\[ r = \sqrt{9 + 16} \]

\[ r = \sqrt{25} \]

\[ r = 5 \]

3. Determining the Position of a Point Relative to the Circle

Given a point \((x_1, y_1)\) and the equation of a circle \((x - h)^2 + (y - k)^2 = r^2\):

3.1. Point Within the Circle

If \( (x_1 - h)^2 + (y_1 - k)^2 < r^2 \), the point is inside the circle.

Example:

For the circle \( (x - 2)^2 + (y + 3)^2 = 25 \) and point \((3, -1)\):

\[ (3 - 2)^2 + (-1 + 3)^2 = 1^2 + 2^2 = 1 + 4 = 5 \]

Since \( 5 < 25 \), the point is inside the circle.

3.2. Point On the Circle

If \( (x_1 - h)^2 + (y_1 - k)^2 = r^2 \), the point is on the circle.

Example:

For the circle \( (x - 2)^2 + (y + 3)^2 = 25 \) and point \((7, -3)\):

\[ (7 - 2)^2 + (-3 + 3)^2 = 5^2 + 0^2 = 25 \]

Since \( 25 = 25 \), the point is on the circle.

3.3. Point Outside the Circle

If \( (x_1 - h)^2 + (y_1 - k)^2 > r^2 \), the point is outside the circle.

Example:

For the circle \( (x - 2)^2 + (y + 3)^2 = 25 \) and point \((10, -1)\):

\[ (10 - 2)^2 + (-1 + 3)^2 = 8^2 + 2^2 = 64 + 4 = 68 \]

Since \( 68 > 25 \), the point is outside the circle.

4. Finding the Center of a Circle

4.1. Finding the Center from the Equation

Given the equation of a circle \((x - h)^2 + (y - k)^2 = r^2\), the center is \((h, k)\).

Example:

For the circle \( (x + 1)^2 + (y - 2)^2 = 16 \),

the center is \((-1, 2)\).

4.2. Finding the Center from Given Coordinates

Given two points on the circle \((x_1, y_1)\) and \((x_2, y_2)\) and the radius \( r \), you can find the center by solving the system of equations formed by the distance formula.

Example:

Given points \((3, 4)\) and \((7, 8)\) on a circle with radius 5, the center \((h, k)\) must satisfy:

\[ (3 - h)^2 + (4 - k)^2 = 25 \]

\[ (7 - h)^2 + (8 - k)^2 = 25 \]

Solving these equations simultaneously will give you the coordinates of the center \((h, k)\).

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