Equations of Circles: Your Complete Study Guide

Hey everyone! Welcome to the study notes for "Equations of Circles". You see circles everywhere, from the wheels of a bus to a pizza. But did you know we can describe these perfect shapes using algebra? That's exactly what this chapter is about!

We'll learn how to write the "address" of any circle on a graph, find out where lines cross circles, and solve some cool problems along the way. Understanding this is super useful, not just for exams, but for things like GPS, computer graphics, and design. Let's get started!


Part 1: The Secret Identity of a Circle - Its Equation

Every circle has a unique equation, just like you have a unique ID card. This equation tells us two very important things: where its centre is and how big its radius is.

1.1 The Standard Form: The Easy-to-Read Version

Think of a circle as a set of points that are all the same distance from a central point. This distance is the radius, and the central point is the centre. To turn this idea into an equation, we just need a little help from a friend you've met before: the distance formula!

Quick Prerequisite Review: The Distance Formula

The distance `d` between two points (x₁, y₁) and (x₂, y₂) is:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Now, let's say a circle has its centre at (h, k) and a radius of r. If we pick any point (x, y) on the edge of the circle, the distance between (x, y) and (h, k) must be `r`. Using the distance formula, we get:

$$\sqrt{(x - h)^2 + (y - k)^2} = r$$

To get rid of the square root, we square both sides. This gives us the Standard Form of the equation of a circle:

$$(x - h)^2 + (y - k)^2 = r^2$$
How to Use the Standard Form:
  • (h, k) is the coordinate of the centre.
  • r is the length of the radius.
Memory Aid & Common Mistake:

Notice the minus signs: (x - h) and (y - k). This means the coordinates of the centre are the opposite of what you see in the brackets!
Example: If the equation is (x - 3)² + (y + 5)² = 16...
- The centre is at (3, -5), NOT (-3, 5).
- The radius is $$\sqrt{16} = 4$$, NOT 16. Don't forget to take the square root!

Let's Try an Example:

Find the equation of a circle with its centre at (2, -1) and a radius of 3.

Step 1: Identify h, k, and r.
Here, h = 2, k = -1, and r = 3.

Step 2: Plug them into the standard form equation: $$(x - h)^2 + (y - k)^2 = r^2$$

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

Step 3: Simplify the equation.

$$(x - 2)^2 + (y + 1)^2 = 9$$
And that's it! You've written the equation of the circle.

Key Takeaway for Standard Form:

The standard form is your best friend for finding the centre and radius quickly. Just remember to watch the signs for the centre and take the square root for the radius.


1.2 The General Form: The Disguised Version

Sometimes, the circle's equation is expanded out and looks a bit messier. This is called the General Form. It's useful, but it hides the centre and radius. Don't worry, we can easily unmask it!

The general form looks like this:

$$x^2 + y^2 + Dx + Ey + F = 0$$

where D, E, and F are just numbers.

Key Features of the General Form:
  • The coefficients of x² and y² are both 1.
  • If they are not 1 but are equal (e.g., 3x² + 3y²...), you must divide the entire equation by that number before you do anything else!

1.3 Finding the Centre and Radius from the General Form

This is a super important skill. How do we find the centre (h, k) and radius r from $$x^2 + y^2 + Dx + Ey + F = 0$$?

Method 1: Completing the Square (The "Proper" Method)

This method converts the general form back into the standard form. It's a reliable process that always works.

Example: Find the centre and radius of the circle $$x^2 + y^2 - 6x + 8y - 11 = 0$$.

Step 1: Group the x-terms and y-terms together, and move the constant (F) to the other side.

$$(x^2 - 6x) + (y^2 + 8y) = 11$$

Step 2: Complete the square for the x-terms and y-terms. Take half of the coefficient of x (or y), square it, and add it to both sides.

  • For x: Half of -6 is -3. (-3)² is 9.
  • For y: Half of 8 is 4. (4)² is 16.

$$(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16$$

Step 3: Factor the perfect squares and simplify the right side.

$$(x - 3)^2 + (y + 4)^2 = 36$$

Step 4: Now it's in standard form! Read off the centre and radius.

  • Centre (h, k): (3, -4)
  • Radius r: $$\sqrt{36} = 6$$
Method 2: Using the Formulas (The Shortcut)

For the equation $$x^2 + y^2 + Dx + Ey + F = 0$$, you can use these handy formulas. They come from the completing the square method.

  • Centre: $$ (-\frac{D}{2}, -\frac{E}{2}) $$
  • Radius: $$ \sqrt{(-\frac{D}{2})^2 + (-\frac{E}{2})^2 - F} $$

Let's use the same example: $$x^2 + y^2 - 6x + 8y - 11 = 0$$

  • Here, D = -6, E = 8, F = -11.
  • Centre: $$ (-\frac{-6}{2}, -\frac{8}{2}) = (3, -4) $$
  • Radius: $$ \sqrt{(3)^2 + (-4)^2 - (-11)} = \sqrt{9 + 16 + 11} = \sqrt{36} = 6 $$

You get the same answer! The shortcut is faster, but make sure you understand where it comes from. And remember, it only works if the coefficients of x² and y² are 1!

Key Takeaway for General Form:

The general form hides the circle's properties. Use either completing the square or the shortcut formulas to convert it back to the standard form to find the centre and radius.


1.4 Finding the Equation from Three Points on the Circle

This is a slightly tougher problem, but it's just about setting up and solving equations. If you're given three points, none of which form a straight line, there is one unique circle that passes through them all.

The Step-by-Step Process:

Example: Find the equation of the circle passing through A(1, 0), B(-1, 2), and C(3, 4).

Step 1: Start with the general form: $$x^2 + y^2 + Dx + Ey + F = 0$$. Our goal is to find D, E, and F.

Step 2: Substitute each point into the equation to create three new equations.

  • For A(1, 0): $$(1)^2 + (0)^2 + D(1) + E(0) + F = 0 \rightarrow 1 + D + F = 0$$ (Equation 1)
  • For B(-1, 2): $$(-1)^2 + (2)^2 + D(-1) + E(2) + F = 0 \rightarrow 5 - D + 2E + F = 0$$ (Equation 2)
  • For C(3, 4): $$(3)^2 + (4)^2 + D(3) + E(4) + F = 0 \rightarrow 25 + 3D + 4E + F = 0$$ (Equation 3)

Step 3: Now you have a system of three linear equations with three unknowns (D, E, F). Solve them simultaneously. (This can take some time, so be careful with your algebra!)

Solving these gives: $$D = -\frac{8}{3}, E = -\frac{14}{3}, F = \frac{5}{3}$$

Step 4: Substitute D, E, and F back into the general form.

$$x^2 + y^2 - \frac{8}{3}x - \frac{14}{3}y + \frac{5}{3} = 0$$

To make it look neater, you can multiply the whole equation by 3:

$$3x^2 + 3y^2 - 8x - 14y + 5 = 0$$

1.5 Is a Point Inside, Outside, or On the Circle?

How can you tell the position of a point relative to a circle? There are two simple ways.

Question: Is the point P(5, 3) inside, outside, or on the circle $$(x - 2)^2 + (y + 1)^2 = 25$$?

Method 1: Compare Distances
  1. Find the circle's centre and radius. Centre is (2, -1), Radius is $$\sqrt{25} = 5$$.
  2. Calculate the distance between the point P(5, 3) and the centre (2, -1).
    $$d = \sqrt{(5 - 2)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
  3. Compare this distance to the radius.
    Here, distance (5) = radius (5).
    Conclusion: The point P is ON the circle.
  • If distance < radius, the point is INSIDE.
  • If distance > radius, the point is OUTSIDE.
Method 2: Substitute and Check
  1. Rearrange the circle equation so it equals zero: $$(x - 2)^2 + (y + 1)^2 - 25 = 0$$.
  2. Substitute the point's coordinates (x=5, y=3) into the left side of the equation.
    $$(5 - 2)^2 + (3 + 1)^2 - 25 = 3^2 + 4^2 - 25 = 9 + 16 - 25 = 0$$
  3. Check the result.
    Here, the result is 0.
    Conclusion: The point P is ON the circle.
  • If the result is < 0, the point is INSIDE.
  • If the result is > 0, the point is OUTSIDE.

Part 2: When Lines and Circles Meet

Imagine a straight road and a circular park. The road could miss the park completely, just touch the edge, or cut right through it. Algebraically, we can find out exactly what happens and where!

2.1 Finding the Points of Intersection

This is a classic simultaneous equations problem. You have one linear equation (the line) and one quadratic equation (the circle). The solution is the point(s) where they meet.

The Step-by-Step Process:

Example: Find the points of intersection between the line $$y = x - 1$$ and the circle $$x^2 + y^2 = 5$$.

Step 1: You have two equations.
(1) $$y = x - 1$$
(2) $$x^2 + y^2 = 5$$

Step 2: Substitute the linear equation into the circle equation. Replace `y` in the circle equation with `(x - 1)`.

$$x^2 + (x - 1)^2 = 5$$

Step 3: Expand and simplify to get a single quadratic equation.

$$x^2 + (x^2 - 2x + 1) = 5$$ $$2x^2 - 2x + 1 = 5$$ $$2x^2 - 2x - 4 = 0$$ $$x^2 - x - 2 = 0$$ (Divided by 2 to make it simpler)

Step 4: Solve this quadratic equation for `x`. We can factor this one!

$$(x - 2)(x + 1) = 0$$

So, $$x = 2$$ or $$x = -1$$.

Step 5: Substitute these x-values back into the linear equation (it's much easier!) to find the corresponding y-values.

  • When x = 2: $$y = 2 - 1 = 1$$. So one point is (2, 1).
  • When x = -1: $$y = -1 - 1 = -2$$. So the other point is (-1, -2).

Conclusion: The line and circle intersect at two points: (2, 1) and (-1, -2).

2.2 Using the Discriminant ($$Δ$$) to Find the Number of Intersections

Sometimes you don't need to know where they intersect, just how many times. This is where the discriminant comes in handy!

Quick Prerequisite Review: The Discriminant

For a quadratic equation $$ax^2 + bx + c = 0$$, the discriminant is $$Δ = b^2 - 4ac$$.

  • If $$Δ > 0$$, there are two distinct real roots.
  • If $$Δ = 0$$, there is one repeated real root.
  • If $$Δ < 0$$, there are no real roots.
Connecting $$Δ$$ to Intersections:

After you substitute the line into the circle and get your quadratic equation, you can use the discriminant to determine the number of intersections without solving it fully.

  • If **$$Δ > 0$$**, there are **2 points of intersection**.
  • If **$$Δ = 0$$**, there is **1 point of intersection** (the line is a tangent).
  • If **$$Δ < 0$$**, there are **0 points of intersection** (they don't touch).
Let's Try an Example:

Find the number of points of intersection of the line $$y = 3x + 10$$ and the circle $$x^2 + y^2 = 10$$.

Step 1: Substitute the line into the circle.

$$x^2 + (3x + 10)^2 = 10$$

Step 2: Expand and simplify to get a standard quadratic equation $$ax^2 + bx + c = 0$$.

$$x^2 + (9x^2 + 60x + 100) = 10$$ $$10x^2 + 60x + 90 = 0$$ $$x^2 + 6x + 9 = 0$$ (Divided by 10)

Step 3: Identify a, b, and c. Here, a = 1, b = 6, c = 9.

Step 4: Calculate the discriminant $$Δ = b^2 - 4ac$$.

$$Δ = (6)^2 - 4(1)(9) = 36 - 36 = 0$$

Conclusion: Since $$Δ = 0$$, the line is a tangent to the circle, and there is exactly one point of intersection.

Did you know?

The condition $$Δ = 0$$ is the key to solving many problems involving tangents to a circle, like finding the equation of a tangent line from a point or finding unknown constants in a line's equation.

Key Takeaway for Intersections:

To find where a line and circle meet, use substitution to create a single quadratic equation. Solve it to find the points, or use the discriminant ($$Δ$$) to quickly find the number of intersections.