Coordinate Geometry of the Circle for the TMUA

Coordinate geometry allows us to describe a circle algebraically. Every point $(x, y)$ on a circle shares a specific relationship based on its distance from the centre.

Defining the Circle at the Origin

Consider a circle with its centre at the origin $(0, 0)$ and a radius of 1. By definition, every point on this circle is exactly 1 unit away from the origin. Using Pythagoras Theorem, we can see that for any point $(x, y)$ on the circumference, the horizontal distance is $x$ and the vertical distance is $y$.

Thus, the equation of the unit circle is $x^2 + y^2 = 1^2$. Points inside the circle satisfy the inequality $x^2 + y^2 < 1^2$, while points outside satisfy $x^2 + y^2 > 1^2$. If we increase the radius to $r$, the equation becomes $x^2 + y^2 = r^2$.

The General Equation in Standard Form

If the centre of the circle is moved from the origin to a general point $(a, b)$, we can find the equation using graph shifting or Pythagoras Theorem. Any point $(x, y)$ on the circle must be a distance $r$ from $(a, b)$.

This leads to the standard form of the circle equation: $(x - a)^2 + (y - b)^2 = r^2$

You must be able to identify the radius and centre immediately from this form. For example:

1. The circle $(x - 2)^2 + (y - 3)^2 = 25$ has its centre at $(2, 3)$ and a radius of 5 (the square root of 25).
2. The circle $(x - 2)^2 + (y + 4)^2 = 18$ has its centre at $(2, -4)$ and a radius of $\sqrt{18} = 3\sqrt{2}$.

The Expanded Form and Completing the Square

Circles are often given in the expanded form: $x^2 + y^2 + cx + dy + e = 0$. To find the centre and radius, we use the method of completing the square for both the $x$ and $y$ terms.

Worked Example 1

Find the centre and radius of $x^2 + y^2 + 4x + 2y - 12 = 0$.

1. Group terms: $(x^2 + 4x) + (y^2 + 2y) - 12 = 0$.
2. Complete the square for $x$: $(x + 2)^2 - 4$.
3. Complete the square for $y$: $(y + 1)^2 - 1$.
4. Combine: $(x + 2)^2 - 4 + (y + 1)^2 - 1 - 12 = 0$.
5. Rearrange: $(x + 2)^2 + (y + 1)^2 = 17$.

The centre is at $(-2, -1)$ and the radius is $\sqrt{17}$.

Worked Example 2: Non Circles

Consider $x^2 + y^2 - 4x - 6y + 20 = 0$. Completing the square gives $(x - 2)^2 - 4 + (y - 3)^2 - 9 + 20 = 0$, which simplifies to $(x - 2)^2 + (y - 3)^2 = -7$. Since the sum of squares cannot be negative, there are no real coordinates that satisfy this equation. It is not a circle.

Worked Example 3: Different Coefficients

Find the centre and radius of $2x^2 + 2y^2 - 4x - 8y - 19 = 0$. First, divide the entire equation by 2 to ensure the $x^2$ and $y^2$ coefficients are 1: $x^2 + y^2 - 2x - 4y - 9.5 = 0$. Completing the square results in $(x - 1)^2 - 1 + (y - 2)^2 - 4 - 9.5 = 0$, or $(x - 1)^2 + (y - 2)^2 = 14.5$. The centre is $(1, 2)$ and the radius is $\sqrt{14.5}$.

Tangents and the Discriminant

A line is tangent to a circle if it intersects it at exactly one point. To solve problems involving tangents, we substitute the line equation into the circle equation and apply the discriminant condition $b^2 - 4ac = 0$ to the resulting quadratic.

Worked Example: Finding Tangent Constants

Find the values of $c$ for which $y = 2x + c$ is tangent to the circle $(x - 3)^2 + (y - 2)^2 = 9$.

1. Substitute: $(x - 3)^2 + (2x + c - 2)^2 = 9$.
2. Expand: $x^2 - 6x + 9 + 4x^2 + 4(c - 2)x + (c - 2)^2 = 9$.
3. Rearrange into a quadratic in $x$: $5x^2 + 2(c - 7)x + (c - 2)^2 = 0$.
4. Apply $b^2 - 4ac = 0$: $4(c - 7)^2 - 20(c - 2)^2 = 0$.
5. Simplify: $(c - 7)^2 = 5(c - 2)^2$. Taking square roots gives $c - 7 = \pm \sqrt{5}(c - 2)$. Solving these linear equations provides the two possible values of $c$.

Shortest Distance Problems

Calculating the shortest distance between a circle and a line is a common coordinate geometry task. The shortest distance usually lies along the line passing through the centre of the circle that is perpendicular to the given line.

Worked Example: Closest Distance

Find the closest distance between $y = x + 11$ and the circle $(x + 2)^2 + (y - 3)^2 = 9$.

1. To simplify, translate the circle to the origin by replacing $x$ with $x - 2$ and $y$ with $y + 3$. The circle becomes $x^2 + y^2 = 3^2$ and the line becomes $y = x + 6$.
2. In this orientation, the point on the line $y = x + 6$ closest to the origin is $Q(-3, 3)$.
3. The distance $OQ = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.
4. Subtract the radius (3) from this distance to find the shortest gap: $3\sqrt{2} - 3$.