Circle equations and tangent lines: the Year 4 test

It is late September 2026 and Year 4 (the fourth year of Italian high school) in the scienze umane courses (the human sciences and economic-social licei) is working on circles and tangent lines, before exponentials arrive in November-December. The class test falls five to eight weeks from now, so between the end of October and mid-November. With only 2 hours of maths a week there is little time to fix the method, and there are two points where students really go wrong: moving from the general equation to centre and radius, and the tangent lines.

The equation of a circle: two forms, one object

A circle is the set of points in the plane that lie at a fixed distance (the radius) from a fixed point (the centre). If the centre is $C(x_0,y_0)$ and the radius is $r$, the definition turns straight into an equation by requiring that the distance of a generic point $(x,y)$ from $C$ be $r$:

This is called the standard form: its advantage is that centre and radius can be read off at a glance. Expanding the squares and tidying up gives the general form:

where $a=-2x_0$, $b=-2y_0$ and $c=x_0^2+y_0^2-r^2$. They are the same curve written in two different ways: the general form is the one that almost always appears in the text of the test, the standard form is the one you need to draw the circle or to reason about centre and radius. The real work of the chapter is learning to move from one to the other without hesitating.

From general form to centre and radius: the critical point

Given $x^2+y^2+ax+by+c=0$, centre and radius come from:

Here lies the most common mistake of the chapter, and it is exactly the point the test insists on: not every equation of the form $x^2+y^2+ax+by+c=0$ represents a real circle. Under the square root of the radius there is a quantity that must be positive:

This is called the reality condition. If the value is positive, you have a genuine circle with a real radius. If it is exactly zero, the equation represents a single point (the “centre”, with zero radius) — a degenerate circle. If it is negative, no real point satisfies the equation: even though it has the right shape, with the coefficients of $x^2$ and $y^2$ equal to 1 and no mixed term $xy$, it represents nothing in the plane.

The reason this step is so often missed is that it looks like a mechanical move — “I apply the formula” — when in fact it is a real question: does this equation describe anything that exists? Skipping it means going on to compute tangents or intersections for a curve that simply isn’t there.

Watch the signs hidden in the coefficients. If the equation is written as $x^2+y^2-6x+4y-3=0$, then $a=-6$ and $b=4$ (not $6$ and $-4$): the centre is

Mixing up the sign of $a$ or $b$ is the second most common error, and it is purely a reading mistake, not a mistake of method.

A circle through three points

If the text gives three points that are not on one line, you find the equation by requiring the circle to pass through each of them in the general form $x^2+y^2+ax+by+c=0$. This gives a linear system of three equations in the three unknowns $a$, $b$, $c$ (linear, because the terms $x^2$ and $y^2$ become known numbers once you substitute the coordinates of the points). There is nothing to guess: it is a system like any other, just with different names for the unknowns.

Position of a line relative to a circle

There are two equivalent methods, and it is worth knowing both because the text sometimes dictates which one to use.

The system method. Put the line and the circle together as a system, substitute, and you get a quadratic equation in one variable. The discriminant $\Delta$ of that equation tells you everything:

• $\Delta >0$ → the line is a secant: two points of intersection;
• $\Delta =0$ → the line is a tangent: one (double) point;
• $\Delta <0$ → the line is external: no intersection.

The distance method. You calculate the distance $d$ from the centre $C$ to the line and compare it with the radius $r$:

• $d<r$ → secant;
• $d=r$ → tangent;
• $d>r$ → external.

If you need to revise the point-to-line distance from scratch, it is covered in linear systems and the line in Year 2 at the linguistic high school. For a circle the distance method is almost always quicker when the equation of the line is already known, because it avoids solving the quadratic; the system method is more direct when you also need the coordinates of the intersection points, not just how many there are.

The tangent line

There are two typical situations, and they are handled differently.

Tangent at a point $P$ on the circle

If $P$ already lies on the circle, the tangent at $P$ is perpendicular to the radius $CP$ at that point — the basic geometric property of a tangent to a circle. The procedure:

1. Find the slope of the radius, $m_{CP}=\frac{y_P-y_C}{x_P-x_C}$.
2. The slope of the tangent is the negative reciprocal: $m_t=-\frac{1}{m_{CP}}$.
3. Write the line through $P$ with that slope.

If the radius is vertical ($x_P=x_C$), the tangent is horizontal, and vice versa: a case you treat separately, with no formulas, simply by reasoning about perpendicularity.

Tangents from an external point

If $P$ is outside the circle, two tangents pass through $P$ (never just one, never none — that is precisely what it means to be an external point). The method is to impose $\Delta =0$ (or $d=r$) on the pencil of lines through $P$:

Put it in a system with the circle (or require that the distance of $C$ from this line equals $r$) and solve for $m$: you get a quadratic equation in $m$ with two solutions, which are the two tangents. As with the parabola, you must check the vertical line through $P$ separately (the pencil written like this does not include it): if that one is also tangent, you have to add it by hand.

Three worked exercises

1. From general form to centre, radius and reality condition

Check whether $x^2+y^2-4x+2y-4=0$ represents a circle, and if so find its centre and radius.

Here $a=-4$, $b=2$, $c=-4$.

Reality condition:

It is positive: this is a genuine circle, with

2. Line-circle position with the distance method

Given the circle with centre $C(1,2)$ and radius $r=5$, determine the position of the line $3x-4y+9=0$ relative to it.

Distance from the centre to the line:

Comparison with the radius: $d=\frac{4}{5}=0.8$, while $r=5$. Since $d<r$, the line is a secant: it meets the circle at two points.

3. Tangent at a point of the circle

Given $x^2+y^2=25$ (centre at the origin, radius 5), find the tangent at the point $P(3,4)$.

First I check that $P$ really lies on the circle: $3^2+4^2=9+16=25$. It does.

The slope of the radius $OP$ is

The tangent has the negative reciprocal slope:

I write the line through $P(3,4)$:

Quick check: the distance of the origin from this line must be 5.

It checks out.

Frequently asked questions

Why isn’t the reality condition “always satisfied” when the coefficients of $x^2$ and $y^2$ are equal to 1? Because that condition (equal coefficients, no $xy$ term) only guarantees that the equation might be a circle, not that it is one: you still have to check that the number under the square root of the radius is positive.

When is the distance method better, and when the system, for tangents? If you only need to know whether a line is tangent, the distance is quicker. If you also need the coordinates of the point of tangency, the system gives them to you directly as the solution.

Does the perpendicular-radius method also work for tangents from an external point? No: that method requires the point to be on the circle already, because it uses the radius $CP$ as a reference. From an external point there is no radius to $P$ ($P$ is not on the curve), so you need the pencil of lines with $\Delta =0$ or $d=r$.

How we work on this in lessons

With 2 hours a week the risk is reaching the test having seen the theory only once. In lessons I insist on the reality condition as an automatic first step every time you start from the general form — first check that the circle exists, then carry on — and on the two techniques for tangents (perpendicular radius, or the pencil with $\Delta =0$), which I have the student choose case by case instead of applying one by rote.

The method you see here — discriminant or distance for the position of a line relative to a curve, $\Delta =0$ for tangency — is the same one already used for the parabola in Year 3 of the science high school, explained with the same approach in the parabola in Year 3: tangents, pencils and problems with a parameter: anyone who met it there will recognise it straight away. For revising point-to-line distance and the link between systems and lines, see linear systems and the line in Year 2. For the physics of the same class, see kinematics in Year 4: equations of motion and graphs.

To get an idea of where you stand, you can try the practice test on the circle: it is calibrated on the Year 3 science high school programme, where the circle is tackled a year earlier than in the human sciences courses, but the content — equation, centre and radius, tangents — is the same. The correct answer to each question is free; the guided step-by-step solution unlocks with your first lesson. The overall picture of the subject is on the maths page; if you’d like help planning your revision before the test, start with the free introductory call.