Home / IGCSE Add Maths 0606 / Coordinate geometry of the circle / Tangents to a circle
Cambridge IGCSE · Topic 8.3
IGCSE Add Maths 0606 / Topic 8 / IGCSE

A tangent meets the radius at a right angle.

A tangent touches a circle at exactly one point, and there it is perpendicular to the radius. That one fact gives the equation of a tangent at a known point. The discriminant from the last lesson does the rest: it tests whether a line is a tangent, and it finds the tangents from a point outside the circle. Pythagoras gives the tangent length.

Topic 8All coordinate geometry of the circle lessons →
The key idea

At the point of contact \(T\) the tangent is perpendicular to the radius \(CT\): find the gradient of \(CT\), take the negative reciprocal and use the line through \(T\). The normal at \(T\) is the line \(CT\) itself, so it passes through the centre. A line is a tangent exactly when substituting it into the circle gives a quadratic with discriminant \(0\); the repeated root is the point of contact. From an external point \(P\), the tangent length satisfies \(PT^{2} = PC^{2} - r^{2}\).

C T r A B P PC tangent at T PA PB PA² = PC² − r², so PA = PB
Fig. 1: the tangent at T is perpendicular to the radius CT. From an external point P there are two tangents, PA and PB, of equal length; each makes a right angle with its radius, so Pythagoras gives PA² = PC² - r²
Section 01

The tangent at a known point.

If you know the point of contact \(T\), you do not need the discriminant at all. The radius \(CT\) and the tangent are perpendicular, so their gradients multiply to \(-1\). The normal at \(T\) is the line perpendicular to the tangent, which is the line \(CT\) itself: every normal to a circle passes through the centre.

Worked example Tangent from an expanded equation
The circle \(x^{2} + y^{2} - 4x + 6y - 12 = 0\) passes through \(T(5, 1)\). Find the equation of the tangent at \(T\).
  1. Complete the square: \((x - 2)^{2} + (y + 3)^{2} = 4 + 9 + 12 = 25\), so \(C(2, -3)\) and \(r = 5\).
  2. Check \(T\): \((5 - 2)^{2} + (1 + 3)^{2} = 9 + 16 = 25\), so \(T\) is on the circle.
  3. Gradient of \(CT = \dfrac{1 - (-3)}{5 - 2} = \dfrac{4}{3}\), so the tangent gradient is \(-\dfrac{3}{4}\).
  4. \(y - 1 = -\dfrac{3}{4}(x - 5) \Rightarrow 4y - 4 = -3x + 15\).
Answer: \(3x + 4y = 19\)
Worked example Normal and tangent
The point \(A(2, 1)\) lies on the circle \((x - 1)^{2} + (y + 2)^{2} = 10\). Find the equations of the normal and the tangent at \(A\).
  1. \(C(1, -2)\). Gradient of \(CA = \dfrac{1 - (-2)}{2 - 1} = 3\).
  2. Normal: gradient 3 through \(A\): \(y - 1 = 3(x - 2)\), so \(y = 3x - 5\). Check: at \(x = 1\), \(y = -2\), the centre.
  3. Tangent: gradient \(-\dfrac{1}{3}\) through \(A\): \(y - 1 = -\dfrac{1}{3}(x - 2) \Rightarrow 3y - 3 = -x + 2\).
Answer: normal \(y = 3x - 5\); tangent \(x + 3y = 5\)
Your turn
Find the equation of the tangent to \(x^{2} + y^{2} = 25\) at the point \((3, -4)\).
Examiner trap

The tangent goes through the point of contact \(T\), not through the centre. Using \(C\) in \(y - y_{1} = m(x - x_{1})\) gives a line parallel to the tangent through the middle of the circle. And 'negative reciprocal' means both: \(\frac{4}{3}\) becomes \(-\frac{3}{4}\), not \(-\frac{4}{3}\).

Section 02

Is this line a tangent?

When the point of contact is not given, use the method from the last lesson. Substitute the line into the circle to get a quadratic in \(x\). Two roots: the line cuts the circle. No real roots: it misses. A repeated root (discriminant \(= 0\)): it touches, so it is a tangent, and the repeated root is the \(x\)-coordinate of the point of contact.

Worked example Show a line is a tangent
Show that the line \(y = 2x + 5\) is a tangent to the circle \(x^{2} + y^{2} = 5\), and find the point of contact.
  1. Substitute: \(x^{2} + (2x + 5)^{2} = 5 \Rightarrow 5x^{2} + 20x + 20 = 0 \Rightarrow x^{2} + 4x + 4 = 0\).
  2. Discriminant: \(4^{2} - 4(1)(4) = 0\), a repeated root, so the line is a tangent.
  3. \((x + 2)^{2} = 0\) gives \(x = -2\), and then \(y = 2(-2) + 5 = 1\).
  4. Check: \((-2)^{2} + 1^{2} = 5\).
Answer: Tangent, touching at \((-2, 1)\)
Worked example Find k for a tangent
Find the values of \(k\) for which the line \(y = x + k\) is a tangent to the circle \((x - 1)^{2} + (y - 2)^{2} = 8\), and find each point of contact.
  1. Substitute: \((x - 1)^{2} + (x + k - 2)^{2} = 8 \Rightarrow 2x^{2} + (2k - 6)x + (k - 2)^{2} - 7 = 0\).
  2. Tangent: \((2k - 6)^{2} - 8\left[(k - 2)^{2} - 7\right] = 0 \Rightarrow -4k^{2} + 8k + 60 = 0\).
  3. \(k^{2} - 2k - 15 = 0 \Rightarrow (k - 5)(k + 3) = 0\), so \(k = 5\) or \(k = -3\).
  4. \(k = 5\): \(2x^{2} + 4x + 2 = 0\), so \(x = -1\) and \(y = 4\).
  5. \(k = -3\): \(2x^{2} - 12x + 18 = 0\), so \(x = 3\) and \(y = 0\).
Answer: \(k = 5\), touching at \((-1, 4)\); \(k = -3\), touching at \((3, 0)\)
Your turn
Show that \(y = 3x - 10\) is a tangent to \(x^{2} + y^{2} = 10\) and find the point of contact.
Examiner trap

The repeated root gives only \(x\). Put it back into the LINE to find \(y\), and state a conclusion: 'discriminant \(= 0\), so the line is a tangent'. A bare '\(= 0\)' with no sentence loses the 'show that' mark.

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The radius \(CT\) of a circle has gradient \(\dfrac{2}{3}\). What is the gradient of the tangent at \(T\)?

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A circle has centre \((1, 2)\) and passes through \(T(4, 6)\). What is the gradient of the tangent at \(T\)?

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A point \(P\) is 13 units from the centre of a circle of radius 5. How long is a tangent from \(P\) to the circle?

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A line is substituted into the equation of a circle and the resulting quadratic has discriminant \(0\). The line

Section 03

Tangents from a point outside.

From a point \(P\) outside the circle there are exactly two tangents, and they have equal length (Fig. 1). The radius to the point of contact meets the tangent at a right angle, so triangle \(PTC\) is right-angled at \(T\) and Pythagoras gives the tangent length without finding \(T\).

Worked example Length of a tangent
The circle \((x - 2)^{2} + (y + 1)^{2} = 36\) and the point \(P(8, 7)\) are given. Find the length of a tangent from \(P\) to the circle.
  1. \(C(2, -1)\), \(r^{2} = 36\).
  2. \(PC^{2} = (8 - 2)^{2} + (7 + 1)^{2} = 36 + 64 = 100\).
  3. \(PT^{2} = PC^{2} - r^{2} = 100 - 36 = 64\).
Answer: \(PT = 8\)
Worked example Two tangents through a point
Find the equations of the two tangents to \(x^{2} + y^{2} = 5\) that pass through \(P(3, 1)\).
  1. A line through \(P\) with gradient \(m\): \(y = mx + c\) with \(c = 1 - 3m\).
  2. Substitute: \(x^{2} + (mx + c)^{2} = 5 \Rightarrow (1 + m^{2})x^{2} + 2mcx + c^{2} - 5 = 0\).
  3. Discriminant \(= 0\): \(4m^{2}c^{2} - 4(1 + m^{2})(c^{2} - 5) = 0\), which simplifies to \(c^{2} = 5(1 + m^{2})\).
  4. \((1 - 3m)^{2} = 5 + 5m^{2} \Rightarrow 4m^{2} - 6m - 4 = 0 \Rightarrow (2m + 1)(m - 2) = 0\).
  5. \(m = 2\): \(y - 1 = 2(x - 3)\). \(m = -\dfrac{1}{2}\): \(y - 1 = -\dfrac{1}{2}(x - 3)\).
Answer: \(y = 2x - 5\) and \(x + 2y = 5\) (touching at \((2, -1)\) and \((1, 2)\))
Your turn
The tangents to \(x^{2} + y^{2} = 10\) at \((3, 1)\) and \((1, -3)\) meet at \(P\). Find \(P\) and the length of each tangent.
Examiner trap

\(PT^{2} = PC^{2} - r^{2}\), not \(PC^{2} + r^{2}\): the hypotenuse of the right-angled triangle is \(PC\), the line to the centre. Also the tangent length is not \(PC - r\); that is the distance from \(P\) to the nearest point of the circle.

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The circle \(x^{2} + y^{2} - 6x + 2y - 15 = 0\) passes through \(A(6, 3)\). Find the equation of the tangent at \(A\).

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Find the values of \(c\) for which the line \(y = 2x + c\) is a tangent to the circle \(x^{2} + y^{2} = 20\).

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Find the length of a tangent from \(P(7, 5)\) to the circle \(x^{2} + y^{2} - 4x - 2y - 4 = 0\).

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Lines through \((0, 5)\) with gradient \(m\) are tangents to \(x^{2} + y^{2} = 5\). Find the possible values of \(m\).

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Six original Cambridge-style questions
A tangent at a given point, a normal and a tangent from an expanded equation, showing a line is a tangent, tangents through the origin, two tangents from an external point and the square formed where two tangents meet.