Coordinate Geometry

Coordinate geometry turns geometric statements into algebra. The circle is the centrepiece at CAPE, and almost every circle question reduces to one of two facts: the distance formula, or a tangent meeting a radius at right angles.

Lines, gradients and distance

For points $A(x_1, y_1)$ and $B(x_2, y_2)$:

$$\text{gradient } m = \frac{y_2 - y_1}{x_2 - x_1}$$ $$\text{distance } AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ $$\text{midpoint} = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$$

For $A(1,2)$ and $B(4,6)$:

$$AB = \sqrt{3^2 + 4^2} = 5 \qquad \text{midpoint} = (2.5,\ 4)$$

The equation of a line through $(x_1, y_1)$ with gradient $m$:

$$y - y_1 = m(x - x_1)$$

Through $(2,3)$ with gradient $-2$: $\;y - 3 = -2(x-2)$, so $y = -2x + 7$.

Parallel and perpendicular

Parallel lines have equal gradients.

Perpendicular lines have gradients whose product is $-1$:

$$m_1 m_2 = -1 \qquad \text{equivalently} \qquad m_2 = -\frac{1}{m_1}$$

So a line perpendicular to $y = 3x + 1$ has gradient $-\frac13$; one perpendicular to $y = -\frac23 x$ has gradient $\frac32$.

The equation of a circle

A circle with centre $(a,b)$ and radius $r$:

$$(x-a)^2 + (y-b)^2 = r^2$$

This is the distance formula in disguise: every point on the circle is exactly $r$ from the centre.

$(x-2)^2 + (y+5)^2 = 16$ has centre $(2, -5)$ and radius $4$.

The general form

Expanding gives the alternative form

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

with

$$\text{centre } (-g, -f) \qquad \text{radius } \sqrt{g^2 + f^2 - c}$$

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.

Comparing, $2g = -6$ so $g = -3$, and $2f = 4$ so $f = 2$.

$$\text{centre} = (3, -2) \qquad r = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$$

Or complete the square, which is safer if the formula slips your mind:

$$(x-3)^2 - 9 + (y+2)^2 - 4 - 12 = 0$$ $$(x-3)^2 + (y+2)^2 = 25$$

Same centre, same radius.

Tangents and normals

The one fact that drives every tangent question:

$$\text{A tangent is perpendicular to the radius at the point of contact.}$$

Find the tangent to $x^2 + y^2 - 6x + 4y - 12 = 0$ at the point $(7, 1)$.

First confirm the point is on the circle: $(7-3)^2 + (1+2)^2 = 16 + 9 = 25$ ✓

Gradient of the radius from $(3,-2)$ to $(7,1)$:

$$m_{\text{rad}} = \frac{1-(-2)}{7-3} = \frac34$$

So the tangent has gradient $-\frac43$:

$$y - 1 = -\tfrac43(x - 7)$$ $$3y - 3 = -4x + 28$$ $$4x + 3y = 31$$

Check at $(7,1)$: $28 + 3 = 31$ ✓

The normal at a point is the line through that point and the centre — for a circle, the normal is simply the extended radius.

Intersections and loci

To find where a line meets a circle, substitute the line into the circle and solve the resulting quadratic. The discriminant then tells you the geometry:

This is how you find a tangent from an external point, or the value of $k$ making $y = 2x + k$ tangent to a given circle: set the discriminant to zero and solve for $k$.

Loci

A locus is the set of points satisfying a condition. Turn the condition into an equation using the distance formula.

"The set of points 3 units from $(1, -4)$" becomes

$$(x-1)^2 + (y+4)^2 = 9$$

"The set of points equidistant from $A$ and $B$" is the perpendicular bisector of $AB$ — found by equating the two distances and simplifying, at which point the squared terms cancel and a straight line remains.