Coordinate Geometry of Lines and Circles

Straight Lines in the $xy$ Plane

The equation of a straight line can be expressed in several forms. The most familiar is $y = mx + c$, where $m$ is the gradient and $c$ is the $y$ intercept. For the TMUA, you must also be fluent with $y - y_1 = m(x - x_1)$, which describes a line with gradient $m$ passing through point $(x_1, y_1)$, and the general form $ax + by + c = 0$.

Understanding the Gradient

The gradient $m$ measures the steepness or the rate of change of $y$ with respect to $x$. It tells us how much $y$ changes for every 1 unit increase in $x$.

Geometrically, the gradient is the tangent of the angle $\theta$ that the line makes with the positive $x$ axis, assuming identical scales on both axes ($m = \tan \theta$). For instance, a line at $45^\circ$ has a gradient of $\tan 45^\circ = 1$.

Parallel and Perpendicular Lines

Two lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are parallel if and only if they have the same steepness, meaning $m_1 = m_2$. They are perpendicular if and only if the product of their gradients is $-1$, or $m_1 m_2 = -1$. This relationship excludes vertical and horizontal lines, which should be treated as special cases ($x = k$ and $y = k$ respectively).

The Coordinate Geometry of the Circle

A circle is the set of all points $(x, y)$ at a fixed distance $r$ (the radius) from a central point $(a, b)$. Using Pythagoras theorem, we can derive the standard equation of a circle.

For a circle centred at the origin $(0, 0)$ with radius $r$, the equation is $x^2 + y^2 = r^2$.

When the centre is shifted to $(a, b)$, the equation becomes $(x - a)^2 + (y - b)^2 = r^2$.

Converting Between Forms

Circles may also appear in the form $x^2 + y^2 + cx + dy + e = 0$. To find the centre and radius, you must complete the square for both $x$ and $y$ terms.

Example 1: Find the centre and radius of $x^2 + y^2 + 4x + 2y - 12 = 0$.

First, group the terms: $(x^2 + 4x) + (y^2 + 2y) - 12 = 0$. Completing the square: $(x + 2)^2 - 4 + (y + 1)^2 - 1 - 12 = 0$. Rearranging: $(x + 2)^2 + (y + 1)^2 = 17$. The centre is $(-2, -1)$ and the radius is $\sqrt{17}$.

Example 2: Not every equation of this form is a circle. Consider $x^2 + y^2 - 4x - 6y + 20 = 0$. Completing the square: $(x - 2)^2 - 4 + (y - 3)^2 - 9 + 20 = 0$, which gives $(x - 2)^2 + (y - 3)^2 = -7$. Since a sum of squares cannot be negative in the real number system, no points satisfy this equation. It is not a circle.

Example 3: For $2x^2 + 2y^2 - 4x - 8y - 19 = 0$, first divide by 2 to get $x^2 + y^2 - 2x - 4y - \frac{19}{2} = 0$. Then complete the square to find the centre $(1, 2)$ and radius $\sqrt{\frac{29}{2}}$.

Problems Involving Lines and Circles

Tangency and Intersections

A line can intersect a circle twice, touch it once (as a tangent), or not at all. To find these points, solve the equations simultaneously. This will lead to a quadratic equation. The discriminant $\Delta = b^2 - 4ac$ determines the nature of the intersection:

1. $\Delta > 0$: Two distinct intersections.
2. $\Delta = 0$: One repeated root. The line is a tangent.
3. $\Delta < 0$: No real intersections.

Worked 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: $(x - 3)^2 + (2x + c - 2)^2 = 9$. Expanding: $x^2 - 6x + 9 + 4x^2 + 4(c - 2)x + (c - 2)^2 = 9$. Simplifying: $5x^2 + 2(c - 7)x + (c - 2)^2 = 0$. Setting the discriminant to zero: $4(c - 7)^2 - 20(c - 2)^2 = 0$. Dividing by 4 and rearranging: $(c - 7)^2 = 5(c - 2)^2$. Taking the square root: $c - 7 = \pm \sqrt{5}(c - 2)$. Solving for $c$ gives the two tangent positions.

Shortest Distance

To find the closest distance between a line and a circle, calculate the distance from the centre of the circle to the line and then subtract the radius.

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

One method is to translate both. Shift the circle centre to $(0, 0)$ by replacing $x$ with $x - 2$ and $y$ with $y + 3$. The line becomes $y + 3 = (x - 2) + 11$, or $y = x + 6$. The circle is $x^2 + y^2 = 3^2$.

The distance from $(0, 0)$ to the point $Q(-3, 3)$ on the line is $\sqrt{(-3)^2 + 3^2} = 3\sqrt{2}$. Since the radius is 3, the shortest distance is $3\sqrt{2} - 3$.

Geometric Circle Properties

You should be able to apply standard circle theorems within the coordinate plane:

1. The perpendicular from the centre to a chord bisects the chord.
2. The tangent at any point is perpendicular to the radius at that point.
3. The angle subtended by an arc at the centre is twice the angle at the circumference.
4. The angle in a semicircle is a right angle.
5. Angles in the same segment are equal.
6. Opposite angles in a cyclic quadrilateral add to $180^\circ$.
7. The alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.