Circle Equations and Tangent Lines
Higher
Use coordinates to describe a circle centred at the origin, check its points and find tangent equations. Connect Pythagoras, perpendicular gradients and simultaneous equations.
Before you start: Pythagoras and Coordinate Distances, Linear Graphs (perpendicular lines), Circle Theorems and Circle Proofs and Simultaneous Equations.
1. Why x² + y² = r² describes a circle
Every point on a circle is the same distance, r, from its centre. When the centre is the origin O(0, 0), a point P(x, y) makes a right-angled triangle with horizontal and vertical side lengths |x| and |y|. Pythagoras gives x² + y² = r². Squaring handles either sign of each coordinate.
Worked example — read a circle equation
For x² + y² = 25, the centre is (0, 0) and the radius is √25 = 5. The four axis intercepts are (5, 0), (−5, 0), (0, 5), (0, −5). The right-hand side is radius squared, not the radius.
For x² + y² = 18, r = √18 = 3√2. For an origin-centred circle of radius 7, write x² + y² = 49. Exact surd simplification is covered in Surds and Exact Calculations.
Use equal physical scales on both axes when drawing a circle. The viewing grid above uses the same number of pixels per unit in both directions.
2. Test points and find missing coordinates
Worked example — is (3, 4) on x² + y² = 25?
Substitute: 3² + 4² = 9 + 16 = 25. It matches, so the point is on the circle. (3, 5) gives 9 + 25 = 34, so it is not on this circle.
For an origin-centred circle, a sum smaller than r² means the point is inside; a larger sum means outside. For example, (1, 2) gives 1 + 4 = 5 < 25, so it lies inside.
Worked example — x = 3 on the circle
3² + y² = 25, so y² = 16. Therefore y = 4 or y = −4. Both (3, 4) and (3, −4) lie on the circle. If the question specifies the lower half, choose −4; if it asks for all points, keep both.
A circle is not a single-valued function y of x across its full width: most x-values have two corresponding y-values. Writing y = √(r² − x²) describes only the upper semicircle.
3. Find a tangent equation
A tangent touches the circle at one point. The radius to that point is perpendicular to the tangent. Use perpendicular gradients to turn that geometric fact into a line equation. The theorem itself is taught in Circle Theorems and Circle Proofs.
Method: check that the point is on the circle → find the radius gradient → take its negative reciprocal → use the point in y = mx + c. If the radius is horizontal or vertical, use the special cases in the next section.
Worked example — tangent at P(3, 4)
The circle is x² + y² = 25, and 3² + 4² = 25 confirms P is on it. The radius from (0, 0) to (3, 4) has gradient 4/3. The tangent gradient is −3/4, since (4/3) × (−3/4) = −1.
Write y = −¾x + c. Substitute (3, 4): 4 = −9/4 + c, so c = 25/4. Thus y = −¾x + 25/4, or equivalently 3x + 4y = 25.
Check the point in the line: 3 × 3 + 4 × 4 = 25. The gradient product is −1, so the line is perpendicular to the radius. Both checks are needed: a perpendicular line elsewhere would not be the required tangent.
Worked example — a point with a negative coordinate
At P(−4, 3) on the same circle, the radius gradient is 3/(−4) = −¾. The tangent gradient is 4/3. In y = (4/3)x + c, substitute P: 3 = −16/3 + c, so c = 25/3. The tangent is y = (4/3)x + 25/3, or −4x + 3y = 25. Its gradient is positive, as expected for a perpendicular to a negative-gradient radius.
4. Tangents at the axis endpoints
The negative-reciprocal method needs defined, nonzero gradients. A horizontal radius has a vertical tangent; a vertical radius has a horizontal tangent. Use the point’s coordinate directly.
At (5, 0), the radius lies along the x-axis and the tangent is x = 5. At (0, 5), the radius lies along the y-axis and the tangent is y = 5. The other endpoints give x = −5 and y = −5.
Do not write y = 5 for the tangent at (5, 0). That horizontal line does not even contain (5, 0). A vertical line has an undefined gradient; it cannot be written in the form y = mx + c.
5. Connect line–circle intersections to equations
A line may cross a circle twice, touch it once, or miss it. Each intersection must satisfy both equations. The graphical meaning is the same as in Quadratic Graphs and Graphical Solutions.
Worked example — intersect y = 3 with x² + y² = 25
Substitute y = 3: x² + 9 = 25, hence x² = 16 and x = ±4. The intersections are (−4, 3) and (4, 3). Both give 16 + 9 = 25.
Compare y = 5: substitution gives x² = 0, so there is one intersection, (0, 5), the tangent point. For y = 6, x² = −11 has no real solution, so the line misses the circle. The full substitution method is already in Simultaneous Equations.
Think Like a Mathematician
Why must a tangent equation pass two checks? A line through the correct point can still cut through the circle; a line with the correct perpendicular gradient can be in the wrong place. Passing through the circle point and being perpendicular to its radius identifies the tangent.
Common Mistakes
Radius squared: x² + y² = 25 has radius 5, not 25. Missing sign: y² = 16 gives ±4 until the position is specified. Wrong gradient: tangent and radius gradients differ. Axis endpoints: handle vertical and horizontal lines directly. Half a circle: the positive square root alone gives only the upper semicircle.
Practice — Try It Yourself
Work on squared paper where a graph is requested. Try each question before opening its answer. Graph readings are estimates: give only the precision your scale supports.
1. Give the centre and radius of x² + y² = 49. Write the equation of an origin-centred circle of radius 6.
Centre (0, 0), radius 7. The second equation is x² + y² = 36. Square the radius when writing the equation, and take the positive square root when reading it.
2. Give the exact radius of x² + y² = 50.
r = √50 = √(25 × 2) = 5√2. The radius is positive.
3. Which of (−3, 4), (2, 2), and (5, 1) lie on x² + y² = 25? Classify the others.
(−3, 4): 9 + 16 = 25, on. (2, 2): 4 + 4 = 8 < 25, inside. (5, 1): 25 + 1 = 26 > 25, outside.
4. On x² + y² = 25, x = −4. Find both y-values, then choose the lower point.
16 + y² = 25 gives y² = 9, so y = ±3. The lower point is (−4, −3).
5. Find the tangent to x² + y² = 25 at (4, 3). Check your result.
The radius gradient is ¾, so the tangent gradient is −4/3. Write y = −(4/3)x + c. At (4, 3), c = 3 + 16/3 = 25/3. Thus 4x + 3y = 25. Check: 4 × 4 + 3 × 3 = 25 and (¾)(−4/3) = −1.
6. Find the tangent to x² + y² = 25 at (−3, −4).
The radius gradient is (−4)/(−3) = 4/3, so the tangent gradient is −¾. Substituting gives −4 = 9/4 + c, hence c = −25/4. The tangent is y = −¾x − 25/4, or 3x + 4y = −25. Check at the point: −9 − 16 = −25.
7. Give the tangents to x² + y² = 25 at (−5, 0) and (0, −5).
At (−5, 0), x = −5, a vertical tangent. At (0, −5), y = −5, a horizontal tangent. Do not divide by zero to find a gradient.
8. Find where y = −3 meets x² + y² = 25. Explain why this line is not a tangent.
x² + 9 = 25, so x = ±4. Intersections: (−4, −3) and (4, −3). There are two distinct crossings, so it is not a tangent.
9. A student says the tangent at (3, 4) is y = (4/3)x. Explain two errors.
Its gradient 4/3 is the radius gradient, not the perpendicular tangent gradient −¾. Also y = (4/3)x passes through the centre: it is the radius line, which continues through two circle points. The correct tangent is 3x + 4y = 25.
Frequently Asked Questions
Does this equation describe every possible circle?
The form x² + y² = r² has its centre at the origin. This GCSE lesson focuses on that required form, rather than a separate course on circles with other centres.
Can a circle have two y-values for one x?
Yes. For example x = 3 on x² + y² = 25 gives y = 4 and y = −4. The upper and lower points are both on the circle.
Do I need to memorise a separate tangent formula?
You can derive the required equation using the perpendicular radius and the point on the circle. This also explains why the method works and when to handle an axis endpoint separately.
Continue Learning
Previous: Graph Translations and Reflections
Apply graph ideas: Rates of Change. Revisit Simultaneous Equations for algebraic intersections, or Circle Theorems and Circle Proofs for geometric circle reasoning.