Mathematics | Equations of circles

Equations of Circles: Your Complete Guide

Hey everyone! Welcome to the study notes for "Equations of Circles". Don't worry, this chapter isn't about drawing perfect circles by hand. Instead, we're going to learn how to describe any circle perfectly using the language of algebra. It’s like giving every circle its own unique address on a map!

Understanding this is super useful, not just for exams, but also in the real world. Think about how GPS pinpoints your location, how game designers create circular areas of effect, or how engineers design gears. They all use the principles we're about to learn. Let's get started!

1. The Basic Equation of a Circle (Standard Form)

Let's start with the most important idea. What is a circle? It's simply all the points that are the exact same distance away from a central point. This fixed distance is the radius ($$r$$), and the central point is the centre ($$(h, k)$$).

Quick Review: The Distance Formula

Remember the distance formula? It's just Pythagoras's Theorem in disguise. The distance between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is:

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

We use this exact idea to build the circle's equation!

Building the Standard Form

Imagine any point $$(x, y)$$ on the circle. The distance between this point and the centre $$(h, k)$$ must always be the radius $$r$$.

Using the distance formula:

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

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

Analogy Time! Imagine a dog on a leash tied to a pole. The pole is the centre (h, k). The leash is the radius (r). The path the dog can walk at the full length of the leash is the circle!

Example:

Find the equation of a circle with its centre at (3, -5) and a radius of 4.

Step-by-step solution:

1. Identify the given information: Centre $$(h, k) = (3, -5)$$ and radius $$r = 4$$.
2. Write down the standard form: $$(x - h)^2 + (y - k)^2 = r^2$$.
3. Substitute the values. Be careful with the signs!
   $$h = 3$$, $$k = -5$$, $$r = 4$$.
4. $$(x - 3)^2 + (y - (-5))^2 = 4^2$$
5. Simplify the equation:
   $$ (x - 3)^2 + (y + 5)^2 = 16 $$

And that’s it! That's the unique "address" for this specific circle.

Key Takeaway

The standard form $$(x - h)^2 + (y - k)^2 = r^2$$ is your best friend. It directly tells you the two most important things about a circle: its centre and its radius. Remember to watch out for the signs inside the brackets!

2. The Other Form: General Equation of a Circle

Sometimes, you'll see the circle equation written in a messy, expanded form. Don't panic! It's called the General Form, and we can easily convert it back to the standard form we love.

The big question is: if you are given the general form, how do you find the centre and radius? The answer is a technique you've seen before: completing the square.

Finding the Centre and Radius from General Form

Let's find the centre and radius for the circle: $$x^2 + y^2 - 6x + 4y - 12 = 0$$

Step-by-step solution:

1. Group the x's and y's: Move the constant term (F) to the right side.
   $$(x^2 - 6x) + (y^2 + 4y) = 12$$
2. Complete the square for x: Take the coefficient of x (-6), divide by 2 (-3), and square it (9). Add this to both sides.
   $$(x^2 - 6x + 9) + (y^2 + 4y) = 12 + 9$$
3. Complete the square for y: Take the coefficient of y (4), divide by 2 (2), and square it (4). Add this to both sides.
   $$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$$
4. Factorise and simplify: The expressions in the brackets are now perfect squares.
   $$(x - 3)^2 + (y + 2)^2 = 25$$
5. Identify the centre and radius: Now it's in standard form!
   Centre $$(h, k) = (3, -2)$$
   Radius $$r^2 = 25 \implies r = \sqrt{25} = 5$$

The Shortcut (Use with care!)

For the general form $$x^2 + y^2 + Dx + Ey + F = 0$$:

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

Important Note: For a real circle to exist, the value under the square root for the radius must be positive! If it's zero, you have a single point. If it's negative, it's not a real circle.

Common Mistakes to Avoid

• Sign Errors: For the centre $$(h, k)$$, remember that the standard form is $$(x-h)^2$$ and $$(y-k)^2$$. So if you see $$(x+3)^2$$, it means $$h = -3$$.
• Forgetting to Square Root: The right side of the standard equation is $$r^2$$, not $$r$$. Always take the square root to find the radius.
• General Form Shortcut: When using the shortcut, remember the negative signs for the centre coordinates $$(-D/2, -E/2)$$.

Key Takeaway

The general form hides the circle's properties, but you can always reveal them by completing the square to get back to the standard form.

3. Finding the Equation from Different Clues

DSE questions often give you clues and ask you to be a detective and find the circle's equation. Here are the common scenarios:

Scenario 1: Given the centre and a point on the circle

Find the equation of a circle with centre (1, 2) that passes through the point (4, 6).

Solution:

1. We have the centre $$(h, k) = (1, 2)$$, but we need the radius $$r$$.
2. The radius is simply the distance between the centre (1, 2) and the point on the circle (4, 6). Use the distance formula!
3. $$ r = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
4. Now we have the centre (1, 2) and radius $$r = 5$$. Put it into standard form:
   $$(x - 1)^2 + (y - 2)^2 = 5^2$$
   $$(x - 1)^2 + (y - 2)^2 = 25$$

Scenario 2: Given three points on the circle

This is the trickiest one, but it's just a matter of systematic work. Don't worry, you can do it!

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

Solution:

1. Start with the general form: $$x^2 + y^2 + Dx + Ey + F = 0$$. Our goal is to find D, E, and F.
2. Since each point is on the circle, it must satisfy the equation. Substitute each point in one by one.
   For A(1, 0): $$1^2 + 0^2 + D(1) + E(0) + F = 0 \implies 1 + D + F = 0$$ ... (i)
   For B(-1, 2): $$(-1)^2 + 2^2 + D(-1) + E(2) + F = 0 \implies 5 - D + 2E + F = 0$$ ... (ii)
   For C(3, 4): $$3^2 + 4^2 + D(3) + E(4) + F = 0 \implies 25 + 3D + 4E + F = 0$$ ... (iii)
3. Now you have a system of three linear equations with three unknowns (D, E, F). Solve them simultaneously. (You can use substitution or elimination).
4. Solving this system (try it!) gives: $$D = -4$$, $$E = -6$$, $$F = 3$$.
5. Substitute D, E, and F back into the general equation:
   $$ x^2 + y^2 - 4x - 6y + 3 = 0 $$

4. Position of a Point Relative to a Circle

How can you tell if a point is inside, outside, or exactly on a circle? It's simple!

Take the standard form $$(x-h)^2 + (y-k)^2 = r^2$$. To check a point $$(x_1, y_1)$$, just calculate the value of $$(x_1-h)^2 + (y_1-k)^2$$ and compare it to $$r^2$$.

• If $$(x_1-h)^2 + (y_1-k)^2 < r^2$$, the point is INSIDE the circle.
• If $$(x_1-h)^2 + (y_1-k)^2 = r^2$$, the point is ON the circle.
• If $$(x_1-h)^2 + (y_1-k)^2 > r^2$$, the point is OUTSIDE the circle.

Example:

Is the point P(5, 4) inside, outside, or on the circle $$(x-2)^2 + (y+1)^2 = 40$$?

Solution:

1. Here, $$r^2 = 40$$.
2. Substitute the point P(5, 4) into the left side of the equation:
   $$(5 - 2)^2 + (4 - (-1))^2 = 3^2 + 5^2 = 9 + 25 = 34$$
3. Compare the result (34) with $$r^2$$ (40).
4. Since $$34 < 40$$, the point P(5, 4) is inside the circle.

5. Intersection of a Line and a Circle (Non-Foundation Topic)

This section is especially important if you're aiming for higher grades. It combines what you know about circles with solving simultaneous equations.

When a straight line meets a circle, there are three possibilities:

1. Two intersections: The line is a secant.
2. One intersection: The line is a tangent (it just touches the circle).
3. No intersections: The line misses the circle completely.

We can find out which case it is and find the exact coordinates of intersection using algebra.

Finding the Points of Intersection

Method: Solve the equations simultaneously.

Find the points of intersection of the line $$y = x - 1$$ and the circle $$x^2 + y^2 = 25$$.

Step-by-step solution:

1. You have a linear equation ($$y = x-1$$) and a circle equation ($$x^2 + y^2 = 25$$).
2. Substitute the linear equation into the circle equation to eliminate one variable (here, we eliminate y).
   $$ x^2 + (x-1)^2 = 25 $$
3. Expand and simplify to get a quadratic equation.
   $$ x^2 + (x^2 - 2x + 1) = 25 $$
   $$ 2x^2 - 2x + 1 - 25 = 0 $$
   $$ 2x^2 - 2x - 24 = 0 $$
   $$ x^2 - x - 12 = 0 $$ (Divided by 2 to simplify)
4. Solve the quadratic equation for x (by factoring, formula, etc.).
   $$ (x-4)(x+3) = 0 $$
   $$ x = 4 $$ or $$ x = -3 $$
5. Find the corresponding y-values by substituting these x-values back into the simple linear equation ($$y = x - 1$$).
   If $$x = 4$$, then $$y = 4 - 1 = 3$$. So one point is (4, 3).
   If $$x = -3$$, then $$y = -3 - 1 = -4$$. So the other point is (-3, -4).

The points of intersection are (4, 3) and (-3, -4). Since there are two points, the line is a secant.

Using the Discriminant ($$\Delta$$)

What if the question only asks for the number of intersections? You don't need to solve all the way! Just use the discriminant of the quadratic equation.

Remember for a quadratic $$ax^2 + bx + c = 0$$, the discriminant is $$\Delta = b^2 - 4ac$$.

After you get the simplified quadratic (like $$x^2 - x - 12 = 0$$ in the example above):

• If $$\Delta > 0 \implies$$ Two distinct real roots \implies 2 points of intersection (Secant)
• If $$\Delta = 0 \implies$$ One repeated real root \implies 1 point of intersection (Tangent)
• If $$\Delta < 0 \implies$$ No real roots \implies 0 points of intersection (Line misses)

Pro Tip

This discriminant method is key for problems that ask you to find an unknown. For example: "Find the value(s) of k such that the line $$y = 2x + k$$ is a tangent to the circle $$x^2 + y^2 = 5$$". For this, you would substitute, get a quadratic in terms of x and k, and then set its discriminant $$\Delta = 0$$ to solve for k.

Key Takeaway

To analyse the intersection of a line and a circle, substitute the linear equation into the circle equation. The resulting quadratic equation holds all the secrets. Solve it for the coordinates, or use its discriminant to find the number of intersections.