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Grade 12 · Analytical Geometry
The Circle & Its Tangent (Grade 12)
MARKING GUIDELINE
Marks
32
Duration
50 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 32
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1A circle with centre (a ; b) and radius r has equation …(1)A)(x − a)² + (y − b)² = r²✓B)(x + a)² + (y + b)² = r²C)(x − a)² + (y − b)² = rD)(x − a)² − (y − b)² = r²Answer: A — That is the formula on the information sheet.
- B — the signs inside the brackets are the OPPOSITE of the centre's coordinates
- C — the right-hand side is r SQUARED
- D — the two squares are ADDED, not subtracted
- 1.2The centre of (x − 3)² + (y + 2)² = 25 is …(1)A)(3 ; −2)✓B)(−3 ; 2)C)(3 ; 2)D)(−3 ; −2)Answer: A — The centre's coordinates are the OPPOSITE of the signs inside the brackets.
- B — both signs have been read the wrong way round
- C — the bracket reads (y + 2), so b = −2
- D — the bracket reads (x − 3), so a = +3
- 1.3The RADIUS of (x − 3)² + (y + 2)² = 25 is …(1)A)5✓B)25C)12,5D)√5Answer: A — r² = 25, so r = 5.
- B — that is r², before the square root
- C — halved 25 instead of square-rooting it
- D — square-rooted 5 instead of 25
- 1.4x² + y² − 6x + 4y − 12 = 0 has centre …(1)A)(3 ; −2)✓B)(−6 ; 4)C)(−3 ; 2)D)(6 ; −4)Answer: A — Completing the square gives (x − 3)² + (y + 2)² = 25.
- B — read the coefficients straight off instead of halving them and reversing the signs
- C — halved the coefficients but did not reverse their signs
- D — reversed the signs but did not halve the coefficients
- 1.5The radius of x² + y² − 6x + 4y − 12 = 0 is …(1)A)5✓B)12C)25D)√12Answer: A — r² = 9 + 4 + 12 = 25, so r = 5.
- B — used the constant 12 without adding the completed squares
- C — that is r², before the square root
- D — square-rooted the constant term only
- 1.6A tangent to a circle at a point P is PERPENDICULAR to …(1)A)the radius drawn to P✓B)any chord through PC)every diameter of the circleD)the x-axisAnswer: A — That is what makes the tangent's gradient the negative reciprocal of the radius's gradient.
- B — a chord through P is not generally perpendicular to the tangent
- C — only the diameter THROUGH P qualifies, and that is the radius extended
- D — the tangent's direction depends on where P lies
- 1.7A circle has centre C(2 ; 1) and P(5 ; 5) lies on it. Its radius is …(1)A)5✓B)7C)25D)√7Answer: A — r = √((5 − 2)² + (5 − 1)²) = √25 = 5.
- B — ADDED the two differences instead of squaring and rooting
- C — forgot to take the square root
- D — subtracted the squares instead of adding them
- 1.8The gradient of the radius from C(2 ; 1) to P(5 ; 5) is …(1)A)43✓B)34C)−34D)−43Answer: A — m = (5 − 1) ÷ (5 − 2) = 4/3.
- B — inverted the formula
- C — that is the TANGENT's gradient
- D — changed the sign as well as inverting
- 1.9So the gradient of the TANGENT at P(5 ; 5) is …(1)A)−34✓B)43C)34D)−43Answer: A — The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of 4/3.
- B — that is the RADIUS's gradient
- C — took the reciprocal but left off the minus
- D — changed the sign but did not take the reciprocal
- 1.10The equation of that tangent at P(5 ; 5) is …(1)A)4y = −3x + 35✓B)4y = 3x + 5C)3y = −4x + 35D)y = −3x + 5Answer: A — y − 5 = −¾(x − 5) gives 4y − 20 = −3x + 15, so 4y = −3x + 35.
- B — used the radius's gradient instead of the tangent's
- C — used −4/3 instead of −3/4
- D — never cleared the fraction, so the constant came out wrong
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[12 MARKS]A circle has the equation x2 + y2 − 4x + 6y − 12 = 0.
- 2.1Determine the coordinates of the centre and the length of the radius by completing the square.(5)(x2 − 4x + 4) + (y2 + 6y + 9) = 12 + 4 + 9 (3)
(x − 2)2 + (y + 3)2 = 25 (1)
Centre (2 ; −3), radius 5 (1) - 2.2Show that the point (5 ; 1) lies on the circle.(2)(5 − 2)2 + (1 + 3)2 = 9 + 16 (1) = 25 ✓ (1)
- 2.3Determine the equation of the tangent to the circle at (5 ; 1).(5)mradius = 1 − (−3)5 − 2 = 43 (2)
The tangent is perpendicular: m = −34 (1)
1 = −34(5) + c gives c = 194 (1)
y = −34x + 194 (1)
Question 3
[10 MARKS]Answer the questions below about the same circle.
- 3.1Determine the coordinates of the point on the circle DIAMETRICALLY opposite (5 ; 1).(3)The centre is the midpoint of the diameter (1)
(2 ; −3) = (5 + x2 ; 1 + y2) (1)
(−1 ; −7) (1) - 3.2Determine whether the point (6 ; 1) lies inside, on, or outside the circle. Justify with a calculation.(4)(6 − 2)2 + (1 + 3)2 = 16 + 16 = 32 (2)
32 > 25 = r2 (1)
The point lies OUTSIDE the circle (1) - 3.3Write down the equation of the circle with the same centre but twice the radius.(3)r = 10, so r2 = 100 (1)
(x − 2)2 + (y + 3)2 = 100 (2)
TOTAL: 32 marks
This question paper consists of 3 questions.