Points and Circles Revision Notes for Leaving Cert Mathematics

Points and Circles

When working with circles in coordinate geometry, you need to understand where a point sits relative to a circle. Any point can be in one of three positions: inside, on, or outside the circle.

Understanding point positions

The position of a point relative to a circle depends entirely on how far that point is from the circle's centre compared to the circle's radius.

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The fundamental principle: A point's position is determined by comparing its distance from the centre with the radius length.

Inside the circle: A point lies inside a circle when its distance from the centre is less than the radius length.

On the circle: A point lies on the circle when its distance from the centre equals the radius length.

Outside the circle: A point lies outside a circle when its distance from the centre is greater than the radius length.

Method 1: Distance comparison method

This method involves calculating the actual distance from the point to the centre and comparing it with the radius. The distance comparison method works for all circle equations and gives you precise distance values.

Steps:

Identify the centre coordinates and radius of the circle
Calculate the distance from the centre to your point using the distance formula
Compare this distance with the radius length

The distance formula between points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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Worked Example: Determining Point Positions

Consider the circle $x^2 + y^2 = 13$ and points $(1, -3)$ and $(3, -2)$ .

The radius length is $\sqrt{13}$ since $r^2 = 13$ .

For point $(1, -3)$ :

Distance from centre $(0, 0)$ = $\sqrt{(1-0)^2 + (-3-0)^2} = \sqrt{1 + 9} = \sqrt{10}$
Since $\sqrt{10} < \sqrt{13}$ , the point $(1, -3)$ is inside the circle

For point $(3, -2)$ :

Distance from centre $(0, 0)$ = $\sqrt{(3-0)^2 + (-2-0)^2} = \sqrt{9 + 4} = \sqrt{13}$
Since $\sqrt{13} = \sqrt{13}$ , the point $(3, -2)$ is on the circle

Method 2: Substitution method

This method is often quicker for circles in standard form. The substitution method involves substituting the point's coordinates directly into the circle equation.

For a circle with equation $x^2 + y^2 = r^2$ :

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Critical Comparison Rules:

If $x^2 + y^2 < r^2$ , the point is inside the circle
If $x^2 + y^2 = r^2$ , the point is on the circle
If $x^2 + y^2 > r^2$ , the point is outside the circle

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Worked Example: Using Substitution Method

Investigate if point $(-3, 4)$ is inside, on, or outside the circle $x^2 + y^2 = 16$ .

Solution: Substituting $x = -3$ and $y = 4$ : $(-3)^2 + (4)^2 = 9 + 16 = 25$

Since $25 > 16$ , the point $(-3, 4)$ is outside the circle.

Key formulas to remember

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Essential Formulas:

Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Standard circle equation: $x^2 + y^2 = r^2$ (centre at origin)
Radius from equation: If $x^2 + y^2 = k$ , then $r = \sqrt{k}$

Exam tips

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Strategic Approaches:

The substitution method is usually faster for circles centred at the origin
Always check your arithmetic carefully when squaring negative numbers
Remember that $\sqrt{a} < \sqrt{b}$ when $a < b$ (for positive values)
Draw a quick sketch if you're unsure about your answer

Remember!

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Key Points to Remember:

A point's position relative to a circle depends on its distance from the centre compared to the radius
Inside: distance < radius (or $x^2 + y^2 < r^2$ )
On the circle: distance = radius (or $x^2 + y^2 = r^2$ )
Outside: distance > radius (or $x^2 + y^2 > r^2$ )
Use the substitution method for quick calculations with standard form circles
The distance method works for all circle equations and gives you the actual distances

Points and Circles (Leaving Cert Mathematics): Revision Notes