Coordinate geometry

Parallel lines share a gradient; perpendicular gradients are negative reciprocals, so a gradient of $\tfrac{3}{4}$ pairs with $-\tfrac{4}{3}$. The point-gradient form is almost always faster than $y = mx + c$ because it needs no second step to find $c$; leave the answer in whichever form the question asks for, and if it does not say, any correct rearrangement scores.

The perpendicular bisector of $AB$ combines them: it passes through the midpoint of $AB$ with gradient $-1/m_{AB}$. It is worth recognising on sight because it is the answer to several differently worded questions — “the set of points equidistant from $A$ and $B$” is exactly that line, and the centre of a circle through two points lies on it.

Papers give the circle in that form, or expanded, and expect you to convert. Expanded form is $x^2 + y^2 + 2gx + 2fy + c = 0$, and you recover the centre and radius by completing the square in $x$ and in $y$ separately:

Centre $(3, -2)$, radius 5. Note the signs: the centre is plus 3 and minus 2, the opposite of what appears in the brackets, and the right-hand side is $r^2$, so the radius is its square root — not 25.

Two other routes to a circle’s equation are set regularly: given the centre and a point on it, the radius is the distance between them; given the two ends of a diameter, the centre is their midpoint and the radius half their distance.

One fact drives nearly every circle question: the tangent at a point is perpendicular to the radius drawn to that point. So the gradient of the tangent at $P$ is the negative reciprocal of the gradient of $CP$, where $C$ is the centre — no calculus needed.

Two companions:

• The perpendicular from the centre to a chord bisects it, which turns chord questions into a right-angled triangle with the radius as hypotenuse: $r^2 = d^2 + (\tfrac{1}{2}\,\text{chord})^2$, where $d$ is the distance from centre to chord.
• The shortest distance from the centre to a line decides the relationship: greater than $r$ and the line misses the circle; equal to $r$ and it is a tangent; less than $r$ and it cuts the circle twice.

The heaviest sub-skill in the section, and mechanically identical every time: substitute the line into the circle, collect into a quadratic in $x$, then either solve it (if the question wants the points) or take the discriminant (if it asks how many intersections, or for a value of $k$ making the line a tangent).

When the line is given as $y = mx + k$ with $k$ unknown, the discriminant becomes a quadratic inequality in $k$ — the technique is on the quadratics page. The alternative route for a tangent, comparing the perpendicular distance from the centre to the line with $r$, is equally valid and often quicker.

1. 1.Radius read as r²

   $(x-3)^2 + (y+2)^2 = 25$ has radius 5. The right-hand side is always the square.

2. 2.Centre signs copied from the brackets

   $(x-a)$ gives centre coordinate $+a$. So $(y+2)^2$ means the $y$-coordinate of the centre is $-2$.

3. 3.Differentiating to find a tangent to a circle

   It works but it is slow and error-prone with implicit differentiation, which is not in P1. Use tangent ⊥ radius: negative reciprocal of the gradient from centre to point.

4. 4.Completing the square in only one variable

   The expanded circle needs it in $x$ and in $y$, with both correction terms carried to the right-hand side.

5. 5.Perpendicular gradient as the reciprocal only

   Negative reciprocal. A gradient of $\tfrac{3}{4}$ gives $-\tfrac{4}{3}$, not $\tfrac{4}{3}$.

6. 6.Solving when the question asked 'how many'

   “Show that the line does not meet the circle” wants the discriminant and a statement that it is negative — not an attempt to find roots that do not exist.