Position of a Point to a Circle

Understanding Point Position Relative to a Circle

If you have a circle and any random point, you must be curious about where that point is, right? Is the point inside the circle, exactly on the edge of the circle, or completely outside the circle?

This concept is really important because in real life we often need to know the position of an object relative to a circular area. For example, is your house still within the signal range of a tower that's circular in shape, or is an airplane's position still within radar surveillance.

By using circle equations and point coordinates, we can determine the position of that point mathematically and accurately.

Power of a Point Concept

To determine the position of a point relative to a circle, we use a concept called power of a point. This is a mathematical way to measure "how far" that point is from the circle.

If we have point $A(x_0, y_0)$ and a circle with general equation $x^2 + y^2 + Dx + Ey + F = 0$, then the power of point $A$ is defined as:

So just substitute the point coordinates into the circle equation, easy right?

Three Possible Point Positions

Based on the power of a point value, there are three possible positions:

1. Point inside the circle occurs when $K_A < 0$. This means the point is located in the area inside the circle.

2. Point on the circle occurs when $K_A = 0$. This means the point is exactly located on the edge or circumference of the circle.

3. Point outside the circle occurs when $K_A > 0$. The point position is located outside the circle area.

If we visualize this, it will look like this:

How to Determine Point Position

The process is quite straightforward. First, we identify the circle equation and coordinates of the point to be checked. Second, substitute the point coordinates into the circle equation to get the power of a point value. Third, look at the sign of the substitution result.

If the result is negative, the point is inside. If zero, the point is on the circle. If positive, the point is outside the circle.

Let's try with a concrete example. Say we have point $A(1, -2)$ and circle $x^2 + y^2 = 25$.

Substitute the point coordinates:

Since the power of point value $K_A = 5$ and for circle $x^2 + y^2 = 25$ which has $r^2 = 25$, then $5 < 25$. So point $A(1, -2)$ is located inside the circle.

For circles in the form $x^2 + y^2 = r^2$, we compare the substitution result with $r^2$. If less than $r^2$, point is inside. If equal to $r^2$, point is on the circle. If greater than $r^2$, point is outside.

Application for General Form

If the circle is in general form $x^2 + y^2 + Dx + Ey + F = 0$, the method is the same. Just substitute the point coordinates into the entire equation and look at the sign of the result.

For example, for point $A(1, -2)$ and circle $x^2 + y^2 - 8x - 2y - 8 = 0$:

Since $K_A = -7 < 0$, then point $A(1, -2)$ is located inside the circle.

Practice Problems

1. Determine the position of point $A(3, 4)$ relative to circle $x^2 + y^2 = 16$.

2. Investigate whether point $B(6, -8)$ lies on circle $x^2 + y^2 = 100$.

3. Determine the position of point $C(2, 1)$ relative to circle $x^2 + y^2 - 4x + 6y - 12 = 0$.

4. A circle has equation $(x-3)^2 + (y+2)^2 = 25$. Determine the position of point $D(7, 2)$ relative to that circle.

Answer Key

1. Solution:

   Substitute coordinates of point $A(3, 4)$ into circle equation $x^2 + y^2 = 16$:

   $$K_A = 3^2 + 4^2 = 9 + 16 = 25$$

   Since $K_A = 25 > 16$, then point $A(3, 4)$ is located outside the circle.

2. Solution:

   Substitute coordinates of point $B(6, -8)$ into circle equation $x^2 + y^2 = 100$:

   $$K_B = 6^2 + (-8)^2 = 36 + 64 = 100$$

   Since $K_B = 100 = 100$, then point $B(6, -8)$ is located exactly on the circle.

3. Solution:

   Substitute coordinates of point $C(2, 1)$ into circle equation $x^2 + y^2 - 4x + 6y - 12 = 0$:

   $$K_C = 2^2 + 1^2 - 4(2) + 6(1) - 12$$

   $$K_C = 4 + 1 - 8 + 6 - 12 = -9$$

   Since $K_C = -9 < 0$, then point $C(2, 1)$ is located inside the circle.

4. Solution:

   For circle $(x-3)^2 + (y+2)^2 = 25$ with center $(3, -2)$ and radius $r = 5$.

   Substitute point $D(7, 2)$:

   $$K_D = (7-3)^2 + (2-(-2))^2 = 4^2 + 4^2$$

   $$K_D = 16 + 16 = 32$$

   Since $K_D = 32 > 25 = r^2$, then point $D(7, 2)$ is located outside the circle.