What is tangent gradient without the negative reciprocal?

The tangent is perpendicular to the radius, so use the negative reciprocal of the radius gradient.

SingaporeAdditional MathematicsSyllabus dot point

How do we find where a line meets a circle and the equation of a tangent to a circle?

Find the intersection of a line and a circle, determine tangency using the discriminant or perpendicular radius, and find tangent equations

A focused answer to the O-Level A-Maths outcome on lines meeting circles. Finding intersection points, deciding tangency via the discriminant or perpendicular radius, and writing the equation of a tangent.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to combine the equation of a line with the equation of a circle: to find where they intersect, to decide whether a line cuts, touches or misses a circle, and to write the equation of a tangent at a given point. This ties together the circle, the straight line and the discriminant in one topic.

The answer

Finding intersection points

To find where a line meets a circle, substitute the line's equation into the circle's equation. This gives a quadratic in one variable whose solutions are the $x$ -coordinates (or $y$ -coordinates) of the intersection points; back-substitute into the line to find the partners.

How many intersections

The discriminant of that quadratic decides the geometry:

\Delta > 0 \Rightarrow \text{two points (the line is a chord)},

\Delta = 0 \Rightarrow \text{one point (the line is a tangent)},

\Delta < 0 \Rightarrow \text{no points (the line misses the circle)}.

So tangency is the condition $\Delta = 0$ , and you can solve it to find a value that makes a line touch a circle.

The tangent at a point of contact

A tangent touches the circle at exactly one point and is perpendicular to the radius drawn to that point. So:

Find the gradient of the radius from the centre to the point of contact.
Take its negative reciprocal for the tangent gradient.
Write the tangent line through the point of contact.

The perpendicular-radius shortcut

Because the radius to the point of contact is perpendicular to the tangent, the perpendicular distance from the centre to a tangent line equals the radius. This gives a second way to test or find tangents without the discriminant.

Worked example

Show that the line $y = 2x + 5$ is a tangent to the circle $x^2 + y^2 = 5$ , and find the point of contact.

Step 1: Substitute the line into the circle

x^2 + (2x + 5)^2 = 5 \Rightarrow x^2 + 4x^2 + 20x + 25 = 5.

Step 2: Form the quadratic

5x^2 + 20x + 20 = 0 \Rightarrow x^2 + 4x + 4 = 0.

Step 3: Check the discriminant

$\Delta = 4^2 - 4(1)(4) = 16 - 16 = 0$ , so there is exactly one solution and the line is a tangent.

Step 4: Find the point of contact

$x^2 + 4x + 4 = (x + 2)^2 = 0$ , so $x = -2$ , and $y = 2(-2) + 5 = 1$ . The point of contact is $(-2, 1)$ .

Examples in context

Example 1. A flight path skimming a no-fly zone. Modelling a circular restricted area and a straight flight path, the discriminant of their combined equation reveals whether the path enters (two crossings), just grazes (tangent) or stays clear, a direct safety calculation.

Example 2. Reflecting off a circular wall. The tangent at the point where a ball strikes a circular cushion is perpendicular to the radius, and the reflection is measured from that tangent, so finding the tangent gradient is the first step in a bounce problem.

Try this

Q1. Find where the line $y = x$ meets the circle $x^2 + y^2 = 8$ . [3 marks]

Cue. $2x^2 = 8$ , so $x = \pm 2$ ; points $(2, 2)$ and $(-2, -2)$ .

Q2. State the tangent gradient to a circle centre $(0,0)$ at the point $(0, 4)$ . [2 marks]

Cue. Radius is vertical, so the tangent is horizontal with gradient $0$ (the line $y = 4$ ).

Q3. Determine whether $y = x + 4$ meets the circle $x^2 + y^2 = 4$ . [3 marks]

Cue. $2x^2 + 8x + 12 = 0$ gives $\Delta = 64 - 96 < 0$ , so the line misses the circle.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksFind the coordinates of the points where the line

y = x + 1

meets the circle

x^2 + y^2 = 25

Show worked answer →

Substitute $y = x + 1$ into the circle: $x^2 + (x + 1)^2 = 25$ .

Expand: $x^2 + x^2 + 2x + 1 = 25$ , so $2x^2 + 2x - 24 = 0$ , that is $x^2 + x - 12 = 0$ .

Factorise: $(x + 4)(x - 3) = 0$ , so $x = -4$ or $x = 3$ .

Then $y = x + 1$ gives $y = -3$ or $y = 4$ . The points are $(-4, -3)$ and $(3, 4)$ .

Markers reward the substitution, the resulting quadratic, and both intersection points.

Original5 marksA circle has centre

(0, 0)

and radius

5

. Find the equation of the tangent to the circle at the point

(3, 4)

Show worked answer →

The radius to $(3, 4)$ has gradient $\dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}$ .

The tangent is perpendicular to the radius, so its gradient is $-\dfrac{3}{4}$ .

Equation through $(3, 4)$ : $y - 4 = -\dfrac{3}{4}(x - 3)$ , which tidies to $3x + 4y = 25$ .

Markers reward the radius gradient, the perpendicular tangent gradient, and the tangent equation through the point of contact.