Study notes

Coordinate Geometry & Proofs

The equation of a circle (centre/radius, completing the square, tangency via the discriminant), intersection of a line and a circle, and light coordinate-geometry proofs using gradients.

Learn it step by step

A circle's equation comes straight from the distance-to-centre idea

A circle with centre $(a, b)$ and radius $r$ is the set of all points exactly distance $r$ from $(a, b)$ , giving the standard equation $(x - a)^{2} + (y - b)^{2} = r^{2}$ . Reading off $a$ , $b$ , and $r$ from an equation already in this form is immediate; the special case with centre at the origin simplifies to $x^{2} + y^{2} = r^{2}$ .

A circle of radius r centred at the origin has equation x^2 + y^2 = r^2.

The radius is the distance from the centre to any point on the circle

If a circle's centre is known but not its radius, and one point on the circle is given, the radius is just the distance between those two points (via the distance formula), and the diameter is twice that. This is how you find the equation of a circle described by its centre plus one point it passes through.

The radius joins the centre to any point on the circle; the diameter, twice the radius, passes straight through the centre.

Completing the square turns the expanded (general) form into the standard form

A circle's equation is often given expanded, as $x^{2} + y^{2} + D x + E y + F = 0$ . Complete the square on the $x$ -terms and the $y$ -terms separately: $x^{2} + D x = (x + \frac{D}{2})^{2} - (\frac{D}{2})^{2}$ , and likewise for $y$ . Collect the leftover constants with $F$ and move them to the right-hand side to reach $(x - a)^{2} + (y - b)^{2} = r^{2}$ , from which the centre and radius can finally be read off.

Finding where a line meets a circle: substitute, then solve the quadratic

Substitute the line's equation for $y$ (or $x$ ) into the circle's equation, which gives a quadratic in the remaining variable. Solving that quadratic gives the $x$ -coordinate(s) of the intersection point(s); substitute each back into the line's equation to get the matching $y$ -coordinate(s).

A line is tangent to a circle exactly when that quadratic has ONE repeated root

Substituting a line into a circle's equation always gives a quadratic $p x^{2} + q x + s = 0$ . If the line crosses the circle at two points, this quadratic has two distinct real roots (discriminant $q^{2} - 4 p s > 0$ ); if the line misses the circle entirely, there are no real roots (discriminant $< 0$ ); if the line is tangent, touching at exactly one point, the quadratic has one repeated root, so the discriminant equals exactly zero. Setting $q^{2} - 4 p s = 0$ is the standard way to find an unknown constant (in the line or the circle) that makes a given line tangent.

Gradient and distance are the two basic tools behind every coordinate proof

The gradient of the segment joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ ; the distance between them is $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ . Almost every coordinate-geometry proof (of collinearity, of a right angle, of a shape's type) reduces to computing one or both of these for the relevant pairs of points and then reasoning from the result.

Equal gradients prove collinearity; gradients multiplying to $-1$ prove a right angle

Three points $A$ , $B$ , $C$ are collinear (all on one straight line) exactly when the gradient of $A B$ equals the gradient of $B C$ (they share point $B$ , so equal gradients means the same line). Two segments are perpendicular exactly when the product of their gradients is $- 1$ (a horizontal line, gradient $0$ , paired with a vertical line, undefined gradient, is the one exception, perpendicular by direct inspection rather than the product rule). This is the light coordinate-geometry-proof thread within this topic; the equation-of-a-circle mechanics above is the main content.

Worked examples

Find the centre and radius of the circle

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

by completing the square.

Group and complete the square on $x$ : $x^{2} - 8 x = (x - 4)^{2} - 16$ .
Complete the square on $y$ : $y^{2} + 2 y = (y + 1)^{2} - 1$ .
Substitute back: $(x - 4)^{2} - 16 + (y + 1)^{2} - 1 + 8 = 0$ , i.e. $(x - 4)^{2} + (y + 1)^{2} - 9 = 0$ .
Rearranged: $(x - 4)^{2} + (y + 1)^{2} = 9$ . Centre $(4, - 1)$ , and $r^{2} = 9$ gives radius $3$ .

Find the points where the line

y = 3 x - 5

meets the circle

x^{2} + y^{2} = 25

Substitute $y = 3 x - 5$ into the circle: $x^{2} + (3 x - 5)^{2} = 25$ .
Expand: $x^{2} + 9 x^{2} - 30 x + 25 = 25$ , so $10 x^{2} - 30 x = 0$ , i.e. $10 x (x - 3) = 0$ , giving $x = 0$ or $x = 3$ .
$x = 0$ : $y = 3 (0) - 5 = - 5$ , point $(0, - 5)$ . $x = 3$ : $y = 3 (3) - 5 = 4$ , point $(3, 4)$ .
Check: $0^{2} + (- 5)^{2} = 25$ and $3^{2} + 4^{2} = 9 + 16 = 25$ , both on the circle.

Find the value(s) of

c

for which the line

y = x + c

is tangent to the circle

x^{2} + y^{2} = 18

Substitute $y = x + c$ : $x^{2} + (x + c)^{2} = 18$ , giving $2 x^{2} + 2 c x + (c^{2} - 18) = 0$ .
For tangency this quadratic must have a repeated root, so its discriminant is zero: $(2 c)^{2} - 4 (2) (c^{2} - 18) = 0$ .
$4 c^{2} - 8 c^{2} + 144 = 0$ , so $- 4 c^{2} + 144 = 0$ , giving $c^{2} = 36$ and $c = 6$ or $c = - 6$ .

Triangle

A B C

has vertices

A (0, 0)

B (2, 4)

C (4, 3)

. Show that triangle

A B C

is right-angled at

B

Gradient of $A B = \frac{4 - 0}{2 - 0} = 2$ .
Gradient of $B C = \frac{3 - 4}{4 - 2} = \frac{- 1}{2} = - \frac{1}{2}$ .
Product of the two gradients: $2 \times (- \frac{1}{2}) = - 1$ , so $A B$ is perpendicular to $B C$ , giving a right angle at $B$ .

Mind map

Coordinate Geometry & Circles

Circle equation

(x-a)^2 + (y-b)^2 = r^2 · centre (a,b), radius r · origin case: x^2 + y^2 = r^2

Completing the square

x^2 + Dx -> (x + D/2)^2 - (D/2)^2 · general form -> standard form

Line meets circle

substitute line into circle equation · solve the resulting quadratic

Tangency

discriminant = 0 · exactly one repeated root

Light coordinate proofs

gradient and distance formulas · collinear: equal gradients · right angle: gradients multiply to -1

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