Grade 12 Maths · Analytical Geometry · Find a tangent to a circle

Find a tangent to a circle Grade 12

Find the gradient of the radius to the point of contact, use the fact that the tangent is perpendicular to it to get the tangent's gradient, then write the tangent's equation (for example a radius gradient of 2 gives a tangent gradient of -12).

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Tangent Analytical Geometry Grade 12Kevinmathscience

Lesson 2: How To Find the Equation of a Circle & Tangent to a Circle | Analytical GeometryMlungisi Nkosi Maths & Science

Analytical Geometry Grade 12: L4 Equation of a Tangent to CircleOnline Maths by Miss Pythagoras

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Question 4.2 · May/June 2025 · Paper 2 · National (DBE) · 4 marks

In the diagram, M is the centre of the circle having equation

(x - 3)^{2} + y^{2} = 25

E (- 1; 3)

and T are points on the circle.

EC is a tangent to the circle at E and cuts the x-axis at D.

E D = \frac{15}{4}

units. MT is produced to meet the tangent at

C (- 7; p)

4.2 Determine the equation of the tangent EC in the form

y = m x + c

.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find a tangent to a circle

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3. Question 4.2 · November 2024 · Paper 2 · National (DBE) · 4 marks
In the diagram, M $(1; 3)$ is the centre of the circle. The circle cuts the $x$ -axis at N. ST is a tangent to the circle at P $(3; 9)$ . R $(d; 1)$ , with $d > 0$ , and L lie on the circle. O and V are the $x$ -intercepts of PL and RL respectively. $Diagram for this question$
4.2 Determine the equation of tangent ST to the circle at P.(4)
Show the memo (4 ticks)Memo
1. ✓
  $m_{MP} = 3$ $m_{MP} = \frac{9 - 3}{3 - 1}$
  $m_{MP} = 3$
  1 mark · Calculate gradient between two points
2. ✓
  $m_{ST} = - \frac{1}{m _{MP}}$ $m_{ST} = - \frac{1}{3}$ [tangent $⊥$ radius]
  1 mark · Find a tangent to a circle
3. ✓
  Substitution of $m_{ST}$ and P $(3; 9)$ into the equation of a line $9 = - \frac{1}{3} (3) + c$ OR $y - 9 = - \frac{1}{3} (x - 3)$
  1 mark · Find the equation of a line
4. ✓
  Equation of tangent ST $c = 10$
  $y = - \frac{1}{3} x + 10$
  1 mark · Find a tangent to a circle
4. Question 4.3.1 · May/June 2023 · Paper 2 · National (DBE) · 2 marks
In the diagram, the equation of the circle centred at N $(- 12; 5)$ is $x^{2} + y^{2} + 24 x - 10 y + 153 = 0$ . The equation of the circle centred at M is $(x + 6)^{2} + (y + 3)^{2} = 25$ . PS and PR are tangents to the circle centred at M at S and R respectively. PR is parallel to the $x$ -axis. K $(- 17; - 5)$ is a point on PS. The straight line joining N and M cuts the smaller circle at T and the larger circle at S. $Diagram for this question$
4.3 Determine the equation of the tangent:
4.3.1 PR(2)
Show the memo (2 ticks)Memo
1. ✓
  y-value of RR $(- 6; - 8)$
  $y_{R} = - 3 - 5 = - 8$
  1 mark · Use the equation of a circle
2. ✓
  Answer $y = - 8$
  1 mark · Find a tangent to a circle
5. Question 4.6 · September 2026 · Paper 2 · Provincial · 2 marks
A circle with centre E passes through the origin. The circle intersects the $x$ -axis at G and the $y$ -axis at H. The tangent to the circle at B cuts the $y$ -axis at A $(0; 8)$ and forms an angle of $45°$ with the $x$ -axis. $Diagram for this question$
4.6 For which values of $t$ will the equation $y = x + t$ be a tangent to the circle?(2)
Show the memo (2 ticks)Memo
1. ✓
  $t = 8$ Tangent at B: $y = x + 8$ , so $t = 8$
  1 mark · Find a tangent to a circle
2. ✓
  $t = 0$ $m_{t a n} = 1$
  $y - 0 = 1 (x - 0)$ [tangent passing through origin]
  $y = x$ , so $t = 0$
  1 mark · Find a tangent to a circle
6. Question 4.2.2 · November 2022 · Paper 2 · National (DBE) · 5 marks
In the diagram, $M (3; - 5)$ is the centre of the circle having PN as its diameter. KL is a tangent to the circle at $N (7; - 2)$ . $Diagram for this question$
4.2 Determine the equation of:
4.2.2 KL in the form $y = m x + c$ (5)
Show the memo (5 ticks)Memo
1. ✓
  Substitution $m_{radius} = \frac{- 5 - ( - 2 )}{3 - 7}$
  1 mark · Calculate gradient between two points
2. ✓
  $m_{radius} = \frac{3}{4}$ $m_{radius} = \frac{- 3}{- 4} = \frac{3}{4}$
  1 mark · Calculate gradient between two points
3. ✓
  $m_{tangent} = - \frac{4}{3}$ $m_{tangent} = - \frac{4}{3}$ [radius $⊥$ tangent]
  1 mark · Find a tangent to a circle
4. ✓
  Substitution of $m$ and N(7 ; −2) $- 2 = - \frac{4}{3} (7) + c$ OR $y - (- 2) = - \frac{4}{3} (x - 7)$
  1 mark · Find the equation of a line
5. ✓
  Equation $c = \frac{22}{3}$
  $y = - \frac{4}{3} x + \frac{22}{3}$
  1 mark · Find a tangent to a circle

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