Coordinate Geometry of the Circle for the ESAT

The Fundamental Equation of a Circle

Consider a circle of radius 1 on the $xy$ plane with its centre at the origin $(0, 0)$. Every point $(x, y)$ sitting on this circle is exactly a distance of 1 from the origin. By applying Pythagoras' theorem to the coordinates of any point on this perimeter, we can establish the relationship between $x$ and $y$. This gives us the equation for a unit circle: $x^2 + y^2 = 1^2$.

Points that satisfy the equation $x^2 + y^2 = 1$ lie on the circle. Points satisfying the inequality $x^2 + y^2 < 1$ lie inside the circle, while those satisfying $x^2 + y^2 > 1$ lie outside the circle. If we increase the radius to $r$, the distance from the origin to any point $(x, y)$ on the circle must be $r$. Again, using Pythagoras' theorem, the equation becomes $x^2 + y^2 = r^2$.

Shifting the Centre of the Circle

To find the equation of a circle with radius $r$ centred at any point $(a, b)$, we can use two different perspectives: geometric derivation or graph transformations.

1. Using Pythagoras' theorem: Any point $(x, y)$ on the circle is a distance $r$ from $(a, b)$. The horizontal distance between the points is $(x - a)$ and the vertical distance is $(y - b)$. Squaring these and summing them gives the squared radius: $(x - a)^2 + (y - b)^2 = r^2$

1. Using graph shifting: Starting with the circle $x^2 + y^2 = r^2$ centred at the origin, we can shift it horizontally by $a$ units and vertically by $b$ units. Using standard transformation rules, we replace $x$ with $(x - a)$ and $y$ with $(y - b)$, resulting in the same general form: $(x - a)^2 + (y - b)^2 = r^2$.

You must be able to quickly identify the centre and radius from this form. For example, in $(x - 2)^2 + (y - 3)^2 = 25$, the centre is $(2, 3)$ and the radius is $\sqrt{25} = 5$. In $(x - 2)^2 + (y + 4)^2 = 18$, the centre is $(2, -4)$ and the radius is $\sqrt{18} = 3\sqrt{2}$. Always remember that the value on the right hand side is $r^2$, not $r$.

Expanded Form and Completing the Square

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

Example 1

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

First, group the terms: $x^2 + 4x + y^2 + 2y - 12 = 0$

Complete the square for $x$ and $y$: $(x + 2)^2 - 4 + (y + 1)^2 - 1 - 12 = 0$

Rearrange into the standard form: $(x + 2)^2 + (y + 1)^2 = 17$

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

Example 2: Non-existent Circles

Find the centre and radius of $x^2 + y^2 - 4x - 6y + 20 = 0$.

Completing the square gives: $(x - 2)^2 - 4 + (y - 3)^2 - 9 + 20 = 0$ $(x - 2)^2 + (y - 3)^2 = -7$

Because the sum of squares on the left hand side cannot be negative, there are no real coordinates $(x, y)$ that satisfy this equation. Thus, it does not represent a circle. An equation of the form $x^2 + y^2 + ax + by + c = 0$ only represents a circle if, after completing the square, the constant on the right hand side is positive.

Example 3: Different Coefficients

Find the centre and radius of $2x^2 + 2y^2 - 4x - 8y - 19 = 0$.

Divide the entire equation by 2 first: $x^2 + y^2 - 2x - 4y - \frac{19}{2} = 0$

Now complete the square: $(x - 1)^2 - 1 + (y - 2)^2 - 4 - \frac{19}{2} = 0$ $(x - 1)^2 + (y - 2)^2 = \frac{29}{2}$

The centre is $(1, 2)$ and the radius is $\sqrt{\frac{29}{2}}$.

Note that for an equation to represent a circle, the coefficients of $x^2$ and $y^2$ must be identical. If they differ, such as in $2x^2 + 3y^2 - 4x - 8y - 19 = 0$, or if the $y^2$ term is negative, such as in $2x^2 - 2y^2 - 4x - 8y - 19 = 0$, the equation describes a different geometric shape.

Tangents to Circles

A line is tangent to a circle if it intersects the circle at exactly one point. To find when a line $y = mx + c$ is tangent to a circle, substitute the line equation into the circle equation to create a quadratic in $x$. Since a tangent has only one point of contact, this quadratic must have a single repeated root, meaning its discriminant must be zero.

Tangent Example

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

Substitute $y = 2x + c$ into the circle equation: $(x - 3)^2 + (2x + c - 2)^2 = 9$ $x^2 - 6x + 9 + 4x^2 + 4x(c - 2) + (c - 2)^2 = 9$ $5x^2 + (4c - 8 - 6)x + (c - 2)^2 = 0$ $5x^2 + 2(2c - 7)x + (c - 2)^2 = 0$

For tangency, set the discriminant $b^2 - 4ac = 0$: $[2(2c - 7)]^2 - 4(5)(c - 2)^2 = 0$ $4(2c - 7)^2 - 20(c - 2)^2 = 0$ $(2c - 7)^2 = 5(c - 2)^2$ $2c - 7 = \pm \sqrt{5}(c - 2)$

Solving for $c$ gives two values, corresponding to the two possible tangent lines with gradient 2 shown in the diagram.

Closest Distance Between a Line and a Circle

To find the closest distance between a line and a circle, it is often helpful to translate the system so the circle is centred at the origin.

Distance Example

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

Translate the centre $(-2, 3)$ to $(0, 0)$ by replacing $x$ with $x - 2$ and $y$ with $y + 3$. The circle becomes $x^2 + y^2 = 9$ (radius 3). The line becomes: $y + 3 = (x - 2) + 11 \rightarrow y = x + 6$

From the origin $(0, 0)$, the point on the line $y = x + 6$ closest to the origin is $Q(-3, 3)$. The distance from the origin to $Q$ is $\sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$. Since the circle has a radius of 3, the shortest distance from the line to the circle is the distance to the origin minus the radius: $3\sqrt{2} - 3$.