Equation of a tangent: worked examples and method

Higher · Grade 8 · Non-calculator

This page sets out the method for equation of a tangent in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Equation of a tangent topic page.

Key facts

$x^{2} + y^{2} = r^{2}$ has centre $(0, 0)$ and radius $r$ (not $r^{2}$ ).
The radius to the point $(a, b)$ has gradient $\frac{b}{a}$ , because it goes from the origin.
A tangent is perpendicular to the radius at the point of contact, so its gradient is $- \frac{a}{b}$ .
Perpendicular gradients multiply to $- 1$ : $\frac{b}{a} \times (- \frac{a}{b}) = - 1$ .
Tangent at $(a, b)$ : $y - b = - \frac{a}{b} (x - a)$ , which tidies to $a x + b y = r^{2}$ .
If the point is on an axis (e.g. $(5, 0)$ ) the tangent is vertical or horizontal: $x = 5$ .

The method, step by step

Method: Radius gradient, then negative reciprocal

When you see 'find the equation of the tangent to $x^{2} + y^{2} = r^{2}$ at the point $(a, b)$ ', do this.

Confirm the point is on the circle: $a^{2} + b^{2} = r^{2}$ .
Gradient of the radius from $(0, 0)$ to $(a, b)$ is $\frac{b}{a}$ .
Gradient of the tangent is the negative reciprocal: $- \frac{a}{b}$ .
Substitute into $y - b = m (x - a)$ .
Rearrange into the form the question asks for ( $y = m x + c$ or $a x + b y = c$ ).

Method: The $a x + b y = r^{2}$ shortcut (and how to check it)

When you want a fast answer or a way to check your first method, do this.

For the tangent at $(a, b)$ on $x^{2} + y^{2} = r^{2}$ , write $a x + b y = r^{2}$ .
Check the point satisfies it: $a \cdot a + b \cdot b = a^{2} + b^{2} = r^{2}$ .
Rearrange to $y = m x + c$ if needed and confirm the gradient is $- \frac{a}{b}$ .
Still show the gradient reasoning in the exam — the shortcut alone may not earn method marks.

How the lesson explains it

On the circle

A point (x, y) on x^2 + y^2 = r^2 satisfies that equation. Check before you start.

Radius gradient

From the origin to (x, y) the gradient is y/x.

Perpendicular

The tangent gradient is the negative reciprocal of the radius gradient.

Point on the line

Use y - y1 = m(x - x1) with the point of contact.

Tidy

Expand and rearrange into y = mx + c if the question asks for that form.

Picture

The radius meets the tangent at right angles on the circle.

Show the point is on the circle

The circle centred at the origin with radius r has equation x^2 + y^2 = r^2. A point (x_1, y_1) lies on it only when x_1^2 + y_1^2 = r^2. That check is the first line of the solution, even when the question says the point is on the circle. Substituting catches a copied sign. For example, (3, 4) lies on x^2 + y^2 = 25 because 3^2 + 4^2 = 9 + 16 = 25. The point (2, -3) lies on x^2 + y^2 = 13 because 2^2 + (-3)^2 = 4 + 9 = 13. The negative coordinate is squared, so it still adds.

The radius to (x_1, y_1) runs from (0, 0) to that point. Its gradient is the change in y divided by the change in x, which is y_1/x_1, as long as x_1 is not zero. A negative y_1 makes a negative gradient, so the radius slopes down as x grows. Write the fraction in lowest terms when you can: (-6)/8 is -3/4. The gradient is a property of the radius, not yet of the tangent.

Method: 1. Substitute the point into x^2+y^2=r^2 before you find a gradient. 2. Point (3,4) lies on x^2+y^2=25 because 3^2+4^2=25. 3. The radius from the origin to (x1,y1) has gradient dfracy1x1.

The tangent is perpendicular to the radius

A tangent is perpendicular to the radius at the point of contact. Two lines are perpendicular when their gradients multiply to -1, so if the radius has gradient m_r the tangent has gradient m_t = -1/m_r. With m_r = y_1/x_1 that is m_t = -x_1/y_1, provided y_1 is not zero. The check is short: y_1/x_1 × (-x_1/y_1) = -1. The x_1 and y_1 cancel, and the minus sign remains. If you forget the minus, the lines are not perpendicular and the equation will be a different chord.

Horizontal and vertical cases are the reason the product form is safer than the gradient form. If the point is (r, 0), the radius is horizontal, the tangent is vertical, and a gradient y_1/x_1 is zero while the tangent gradient would require division by zero. The tangent is the line x = r. If the point is (0, r), the radius is vertical and the tangent is the horizontal line y = r. You still do the perpendicular test when both gradients exist. When one of them does not exist, name the horizontal or vertical line instead of forcing a fraction.

Method: 1. The tangent gradient is the negative reciprocal of the radius gradient. 2. If the radius gradient is dfracy1x1, the tangent gradient is -dfracx1y1. 3. Check the product of the gradients is -1.

Building the tangent equation

When both gradients exist, start from the point-gradient equation of the tangent: y - y_1 = -x_1/y_1(x - x_1). Multiply both sides by y_1 to clear the denominator: y_1(y - y_1) = -x_1(x - x_1). Expand: y y_1 - y_1^2 = -x x_1 + x_1^2. Add x x_1 to both sides and add y_1^2 to both sides: x x_1 + y y_1 = x_1^2 + y_1^2. The point is on the circle, so the right-hand side is r^2. The tangent is x x_1 + y y_1 = r^2.

That equation is worth memorising, but only together with the perpendicular reason. The coefficients of x and y are the coordinates of the point, signs included. At (2, -3) on x^2 + y^2 = 13 the tangent is 2x + (-3)y = 13, which tidies to 2x - 3y = 13. At (3, 4) on x^2 + y^2 = 25 it is 3x + 4y = 25. Substitute the point back in at the end. You should recover r^2 on both sides. If you do not, a sign slipped when you copied y_1.

The product form still works for the horizontal and vertical tangents that broke the gradient. At (0, 5) on x^2 + y^2 = 25 it gives y × 5 = 25, so y = 5. At (5, 0) it gives 5x = 25, so x = 5. One equation covers every point of contact on an origin-centred circle. Clear a common factor at the end if the question wants the smallest whole-number coefficients. Dividing 8x - 6y = 100 by 2 gives 4x - 3y = 50, which is the same line.

Method: 1. Start from y-y1=-dfracx1y1(x-x1) and clear the fraction. 2. Rearrange to xx1+yy1=r^2. 3. At (3,4) on x^2+y^2=25, the tangent is 3x+4y=25.

Wording and answer forms in exam questions

A typical question says "the circle with equation x^2 + y^2 = 85 passes through the point P(9, -2). Find an equation of the tangent to the circle at P." Some versions ask you to show the point is on the circle first, and some give the answer form: "in the form y = mx + c" or "in the form ax + by + c = 0 where a, b and c are integers". Read the form before you start, because the last line of working is a rearrangement into that form.

For a circle centred at the origin there is a useful check. The tangent at (a, b) on x^2 + y^2 = r^2 is ax + by = r^2. You still need to show the method in the exam, but the check catches arithmetic slips. Marks are given for the gradient of the radius, the negative reciprocal, a correct straight-line equation, and the final form.

Method: 1. Check that the point satisfies the circle equation, if it is not already given. 2. Find the gradient of the radius from the centre to the point. 3. Take the negative reciprocal for the gradient of the tangent. 4. Use y - y_1 = m(x - x_1) and rearrange into the form asked for.

Tip: If the answer has fractions and the question wants integers, multiply the whole equation through by the denominator.

Finding the point of contact first

Some questions give only one coordinate. For example "P has x-coordinate 3 and lies below the x-axis on the circle x^2 + y^2 = 45". Substitute x = 3 to get 9 + y^2 = 45, so y^2 = 36 and y = ± 6. Below the axis means y is negative, so P = (3, -6). Missing this word "below" or "above" is a common way to reach a wrong tangent.

Another style gives a diagram with a labelled point and says "find the coordinates of P" in an earlier part. Always write the pair in brackets before you use it. Check by substitution that x^2 + y^2 equals the radius squared. Then the gradient from the centre to the point is y/x for a circle centred at the origin.

Method: 1. Substitute the known coordinate into the circle equation. 2. Solve for the missing coordinate and use the clue "above" or "below" to choose the sign. 3. Write the point as a pair and check it in the circle equation. 4. Carry on with the gradient and the tangent.

Circles with a centre away from the origin

When the circle is not centred at the origin, the radius gradient comes from the two points: the centre (p, q) and the point of contact (x_1, y_1). The gradient is (y_1 - q)/(x_1 - p). For a centre (2, 3) and a point (4, 7) it is (7 - 3)/(4 - 2) = 2. Using the gradient from the origin to the point is the standard error, so write the gradient formula with the centre in it.

Then the tangent gradient is the negative reciprocal, here -tfrac12, and the tangent passes through the point of contact (4, 7). The equation is y - 7 = -tfrac12(x - 4), which simplifies to x + 2y - 18 = 0. You do not need the radius for the tangent, but you may need it to check that the point is really on the circle.

Method: 1. Write the centre and the point of contact clearly. 2. Work out the radius gradient as change in y over change in x. 3. Take the negative reciprocal and substitute into y - y_1 = m(x - x_1). 4. Rearrange to the required form and check by putting the point back in.

Using the tangent: axes, areas and conditions for touching

Later parts build on the tangent. To find where it crosses the axes, put x = 0 for the y-intercept and y = 0 for the x-intercept. The tangent and the two axes make a right-angled triangle, so the area is half the product of the intercepts. Two tangents at different points meet where you solve their equations as simultaneous equations.

A different style does not give a point. It gives a line such as y = -2x + k and says it is a tangent to a circle, and asks for k. Substitute the line into the circle, which gives a quadratic in x. A tangent touches once, so the quadratic has exactly one solution and its discriminant b^2 - 4ac is zero. This is a Grade 8 to 9 skill and it often appears with two possible answers for k.

Method: 1. Put x = 0 and y = 0 into the tangent to find the intercepts. 2. Use half base times height for the triangle with the axes. 3. To find where two tangents meet, solve the two equations together. 4. For an unknown k, substitute, set the discriminant equal to zero, and solve.

Remember

Radius gradient is y/x.
Tangent gradient is the negative reciprocal.
Put the point of contact into y - y1 = m(x - x1).

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