- Find the slope and the angle of inclination of a line and write its equation in the form that suits the data
- Recognise parallel and perpendicular lines from their equations and compute the distance from a point to a line and between parallel lines
- Convert a circle between the general and the standard form and find its centre and radius
- Decide the position of a line and a circle and of two circles by comparing distances, and find tangent lines and chord lengths
A radar station at the origin sees everything within 10 km. A ship sails along the straight course 3x + 4y − 40 = 0 (units: km). Will the radar see it, and along what stretch? No drawing is needed: the distance from the station to the course is |−40|/√(3² + 4²) = 8 km, which is less than 10 km, so the ship crosses the radar zone, and the part of its course inside the zone is 2√(10² − 8²) = 12 km long. A line, a circle, a distance and a chord: that is this whole lesson in one example. The official CSCA syllabus has the line “Plane analytic geometry: lines, circles, ellipses, hyperbolas, parabolas”. Lines and circles come first, because the curves of the lessons “The ellipse”, “The hyperbola” and “The parabola” are built with the same tools. The vector approach to angles and perpendicularity is in the lesson “Plane vectors”.
Slope, angle of inclination and the equation of a line
Every line that is not vertical rises or falls at a constant rate. Two numbers describe this rate: the angle between the line and the x-axis, and the slope, which is the tangent of that angle. Chinese textbooks, and therefore CSCA, write the slope as k: y = kx + b; many English textbooks write m: y = mx + c. It is the same number.
倾斜角)The angle α through which the positive direction of the x-axis must be turned anticlockwise until it lies along the line. By convention 0° ≤ α < 180°: a horizontal line has α = 0°, a vertical line α = 90°.
斜率)For a line that is not vertical, k = tan α. The slope shows how much y changes when x grows by 1. A vertical line x = a has an angle of inclination of 90° but no slope, because tan 90° does not exist.
- kslope
- αangle of inclination, 0° ≤ α < 180°, α ≠ 90°
- (x₁, y₁), (x₂, y₂)two points of the line, x₁ ≠ x₂
k > 0: the line rises (0° < α < 90°); k < 0: it falls (90° < α < 180°); k = 0: it is horizontal. If x₁ = x₂, the line is vertical and has no slope.
| α | k = tan α | α | k = tan α |
|---|---|---|---|
| 0° | 0 | 90° | none |
| 30° | √3/3 | 120° | −√3 |
| 45° | 1 | 135° | −1 |
| 60° | √3 | 150° | −√3/3 |
1) Find the slope and the angle of inclination of the line through A(−1, 4) and B(2, 1).
2) The points A(1, 3), B(2, 5) and C(4, m) lie on one line. Find m.
3) What are the slope and the angle of inclination of the line through M(3, −2) and N(3, 5)?
Show solutionHide solution
2) Points on one line give equal slopes: kAB = (5 − 3)/(2 − 1) = 2, so (m − 3)/(4 − 1) = 2 and m = 9.
3) x₁ = x₂ = 3: this is the vertical line x = 3, so α = 90° and the slope does not exist (the formula would divide by zero).
What is the angle of inclination of the line x − √3y + 1 = 0?
A) 30° B) 60° C) 120° D) 150°
Show solutionHide solution
B) 60° belongs to k = √3: the fraction was read upside down. D) 150° belongs to k = −√3/3: a sign was lost when x was moved to the other side. C) combines both mistakes. Faster (see the next section): k = −A/B = −1/(−√3) = √3/3.
Five forms of the equation of a line. A line is fixed by a point and a slope, by two points, or by its intercepts on the axes. Each kind of data has its own form of the equation, and every form can be rewritten as the general form Ax + By + C = 0 — the form in which CSCA usually writes its options.
| Form | Equation | Given | Cannot describe |
|---|---|---|---|
point-slope (点斜式) | y − y₀ = k(x − x₀) | point (x₀, y₀) and k | vertical lines |
slope-intercept (斜截式) | y = kx + b | k and the y-intercept b | vertical lines |
two-point (两点式) | (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁) | two points | lines parallel to an axis |
intercept (截距式) | x/a + y/b = 1 | intercepts a ≠ 0 and b ≠ 0 | lines parallel to an axis and lines through the origin |
general (一般式) | Ax + By + C = 0 | A and B not both 0 | describes every line |
- A, B, Ccoefficients; A and B are not both zero
- kslope
- bthe y-intercept: the line meets the y-axis at (0, b)
An intercept is a coordinate, not a length, so it can be negative. The x-intercept of Ax + By + C = 0 is −C/A (A ≠ 0). If B = 0, the line is vertical: x = −C/A.
Write each line in general form.
1) The line through P(2, −1) with slope 3.
2) The line through A(1, 2) and B(3, −2).
3) The line with x-intercept 3 and y-intercept −2.
Show solutionHide solution
2) k = (−2 − 2)/(3 − 1) = −2; y − 2 = −2(x − 1), so 2x + y − 4 = 0. Check with B: 6 − 2 − 4 = 0 ✓.
3) Intercept form: x/3 + y/(−2) = 1. Multiply by 6: 2x − 3y = 6, so 2x − 3y − 6 = 0.
A line passes through P(2, 3), and its intercepts on the two axes are equal. What is its equation?
A) x + y − 5 = 0 B) 3x − 2y = 0 C) x + y − 5 = 0 or 3x − 2y = 0 D) x − y + 1 = 0
Show solutionHide solution
A is the answer of a student who used only the intercept form, which cannot describe a line through the origin. D has intercepts −1 and 1: equal in length, but not equal.
Parallel and perpendicular lines; distances
- k₁, k₂slopes of the lines y = k₁x + b₁ and y = k₂x + b₂
- b₁, b₂their y-intercepts
Perpendicular slopes are negative reciprocals of each other: 2 and −1/2, −3/4 and 4/3. The test needs both slopes to exist: a vertical line is perpendicular to every horizontal one, although it has no slope.
- A₁, B₁, A₂, B₂coefficients of l₁: A₁x + B₁y + C₁ = 0 and l₂: A₂x + B₂y + C₂ = 0
These conditions work for all lines, vertical ones included, so they are the safest choice in parameter questions. A parallel line keeps A and B and changes only C; a perpendicular line swaps A and B and changes one sign: Ax + By + C = 0 → Bx − Ay + C′ = 0.
1) Find the line through (1, −2) parallel to 3x − y + 4 = 0.
2) Find the line through (2, 3) perpendicular to 2x + y − 1 = 0.
3) For which a are the lines ax + 2y + 1 = 0 and x + (a + 1)y − 2 = 0 perpendicular?
Show solutionHide solution
2) Swap A and B and change one sign: x − 2y + C = 0. The point gives 2 − 6 + C = 0, C = 4: x − 2y + 4 = 0. Check: the slopes −2 and 1/2 multiply to −1 ✓.
3) A₁A₂ + B₁B₂ = a · 1 + 2(a + 1) = 3a + 2 = 0, so a = −2/3.
The lines x + ay + 6 = 0 and (a − 2)x + 3y + 2a = 0 are parallel. What is a?
A) −1 B) 3 C) −1 or 3 D) 1
Show solutionHide solution
Now check for coincidence. At a = 3 both equations become x + 3y + 6 = 0: the same line, so a = 3 is rejected. At a = −1: x − y + 6 = 0 and −3x + 3y − 2 = 0, i.e. x − y + 2/3 = 0 — parallel. Answer A.
C is the trap: it stops after the quadratic equation. B is the coincident case itself.
- (x₀, y₀)the point
- A, B, Ccoefficients of the general equation of the line
- C₁, C₂free terms of the parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, which must have the same A and B
The distance is the length of the perpendicular dropped from the point to the line. The second formula is the first one applied to a point of one of the lines, so first make A and B of both equations equal.
1) Find the distance from (2, −1) to the line 3x + 4y + 8 = 0.
2) Find the distance between the lines 2x − y + 3 = 0 and 4x − 2y − 1 = 0.
3) The distance from the origin to the line x + y + c = 0 is √2. Find c.
Show solutionHide solution
2) Divide the second equation by 2: 2x − y − 1/2 = 0. Then d = |3 − (−1/2)|/√(4 + 1) = (7/2)/√5 = 7√5/10.
3) |c|/√2 = √2, so |c| = 2 and c = ±2: two lines, one on each side of the origin.
The equation of a circle
圆)The set of all points of the plane that are at the same distance r (the radius, 半径) from a fixed point C(a, b) (the centre, 圆心).
- (a, b)centre
- rradius, r > 0
- (x, y)any point of the circle
This is the distance formula |PC| = √((x − a)² + (y − b)²) = r, squared. A point P lies inside, on or outside the circle when |PC| is less than, equal to or greater than r. Some English books write the centre as (h, k); CSCA and Chinese textbooks use (a, b).
- D, E, Fcoefficients of the general equation
It is the standard equation with the brackets opened. The coefficients of x² and y² must be equal, and there must be no xy term. If D² + E² − 4F = 0, the equation describes a single point; if it is negative, no point at all.
1) Write the equation of the circle with centre (−1, 2) that passes through (2, 6).
2) Find the centre and radius of x² + y² − 6x + 2y + 6 = 0.
3) For which m does x² + y² + 2x − 4y + m = 0 describe a circle?
4) Find the circle through O(0, 0), A(4, 0) and B(0, 2).
Show solutionHide solution
2) Complete the squares: (x − 3)² + (y + 1)² = 9 + 1 − 6 = 4. Centre (3, −1), r = 2.
3) D² + E² − 4F = 4 + 16 − 4m > 0, so m < 5.
4) O gives F = 0; A gives 16 + 4D = 0, D = −4; B gives 4 + 2E = 0, E = −2. So x² + y² − 4x − 2y = 0, centre (2, 1), r = √5. Faster: ∠AOB = 90°, so AB is a diameter and its midpoint (2, 1) is the centre.
A line and a circle; two circles
Whether a line cuts a circle is decided by one number: the distance d from the centre to the line. Solving the system of the two equations and looking at the discriminant gives the same answer, but takes longer.
- ddistance from the centre (a, b) to the line: d = |Aa + Bb + C|/√(A² + B²)
- rradius of the circle
In Chinese the three cases are 相交 (intersect), 相切 (tangent) and 相离 (separate). The algebraic check gives the same result: substituting the line into the circle gives a quadratic equation, and Δ > 0, Δ = 0 and Δ < 0 correspond to the three cases.
- |AB|length of the chord
- ddistance from the centre to the line
The perpendicular from the centre to a chord bisects it, so r, d and half the chord form a right triangle (see the drawing).
1) Does the line 3x + 4y − 5 = 0 cut the circle (x − 1)² + (y − 1)² = 4, touch it or miss it?
2) Find the length of the chord it cuts off.
3) For which b is the line y = x + b tangent to the circle x² + y² = 8?
Show solutionHide solution
2) |AB| = 2√(4 − 4/25) = 2√(96/25) = 2 · 4√6/5 = 8√6/5.
3) Write the line as x − y + b = 0. d = |b|/√2 must equal r = 2√2, so |b| = 4 and b = ±4.
- (x₀, y₀)the point of contact on the circle (first formula: centre at O)
- |PT|length of the tangent segment from an outside point P to the point of contact T
- |PC|distance from P to the centre
The first two formulas give the tangent at a point of the circle (“replace one x by x₀ and one y by y₀”). They work because the tangent is perpendicular to the radius at the point of contact; the same right angle gives |PT| by Pythagoras. From a point outside the circle there are exactly two tangents, from a point on it one, from a point inside none.
- 1Check the position
Compare |PC| with r. From a point outside there are two tangents; for a point on the circle use the formula above.
- 2Test the vertical line first
Find the distance from the centre to x = x₀. If it equals r, this vertical line is one of the tangents — the slope method cannot find it.
- 3Use the slope form
Write y − y₀ = k(x − x₀) as kx − y + y₀ − kx₀ = 0 and require d = r.
- 4Solve for k
The equation for k has two roots, or only one when the vertical line is the second tangent. Write both lines in general form.
1) Find the tangent to the circle x² + y² = 25 at the point (3, −4).
2) Find the tangents from P(3, 2) to the circle (x − 1)² + y² = 4 and the length of the tangent segment.
Show solutionHide solution
2) |PC|² = (3 − 1)² + 2² = 8 > 4: P is outside, so there are two tangents. The vertical line x = 3 is at distance |3 − 1| = 2 = r from the centre, so x = 3 is a tangent. Slope form: kx − y + 2 − 3k = 0 and d = |2 − 2k|/√(k² + 1) = 2, so 4 − 8k + 4k² = 4k² + 4 and k = 0: y = 2. Only one k appeared; the vertical line is the second tangent. Tangent segment: √(8 − 4) = 2.
| Position | Condition | Common points | Common tangents |
|---|---|---|---|
outside each other (外离) | d > R + r | 0 | 4 |
externally tangent (外切) | d = R + r | 1 | 3 |
intersecting (相交) | |R − r| < d < R + r | 2 | 2 |
internally tangent (内切) | d = |R − r| > 0 | 1 | 1 |
one inside the other (内含) | d < |R − r| | 0 | 0 |
- D₁, E₁, F₁; D₂, E₂, F₂coefficients of the general equations of the two circles
If two circles intersect, subtracting their general equations removes x² and y² and leaves a line through both common points: the line of the common chord (公共弦).
1) How are the circles x² + y² = 36 and (x − 3)² + (y + 4)² = 1 positioned, and how many common tangents do they have?
2) Find the line of the common chord of x² + y² − 4 = 0 and x² + y² − 4x + 4y − 4 = 0, and the length of the chord.
Show solutionHide solution
2) Subtract: (x² + y² − 4) − (x² + y² − 4x + 4y − 4) = 4x − 4y = 0, so the common chord lies on y = x. This line passes through the centre O of the first circle, so the chord is a diameter of that circle: length 4.
For every real k, the line y = kx + 1 and the circle x² + y² − 2x − 3 = 0:
A) are tangent B) intersect at two points C) have no common points D) intersect or touch, depending on k
Show solutionHide solution
D is what a hurried discriminant with a sign error suggests; checking the fixed point takes ten seconds and needs no algebra.
How CSCA asks about this
- Slope and angle: read k = −A/B from the general equation and use the table of special angles; a range of slopes gives a range of angles that may split at 90°.
- Equations of lines: use the point-slope form for a point and a slope; keep A and B for a parallel line, swap them and change a sign for a perpendicular one; look for the lost case (a vertical line, a line through the origin).
- Parameters: A₁A₂ + B₁B₂ = 0 for perpendicular lines; A₁B₂ − A₂B₁ = 0 for parallel ones, then reject the value that makes the lines coincide.
- Distances: the point-to-line formula; for parallel lines make A and B equal first. Distance questions with a parameter usually have two answers (±).
- Circles: complete the squares (the centre is (−D/2, −E/2)), check D² + E² − 4F > 0 and use geometry: a right angle inscribed in a circle stands on a diameter.
- Line and circle, two circles: compare d with r (or with R + r and |R − r|); chord 2√(r² − d²); tangent at a point x₀x + y₀y = r²; a fixed point inside the circle means “always intersect”, and the shortest chord through it is perpendicular to the radius through that point.
- Time-savers (about 75 s per item): you rarely need to solve a system — d versus r and substituting the options are faster. When the options are equations of lines, first check which ones pass through the given point: this often leaves one option.
The test can be taken in English or in Chinese. These are the terms of this lesson in both languages:
| Term | 中文 | Pinyin |
|---|---|---|
| slope | 斜率 | xiélǜ |
| angle of inclination | 倾斜角 | qīngxiéjiǎo |
| intercept | 截距 | jiéjù |
| general equation of a line | 直线的一般式方程 | zhíxiàn de yībānshì fāngchéng |
| parallel | 平行 | píngxíng |
| perpendicular | 垂直 | chuízhí |
| distance from a point to a line | 点到直线的距离 | diǎn dào zhíxiàn de jùlí |
| circle | 圆 | yuán |
| centre (center) | 圆心 | yuánxīn |
| radius | 半径 | bànjìng |
| chord | 弦 | xián |
| tangent line | 切线 | qiēxiàn |
| intersect / tangent / separate | 相交 / 相切 / 相离 | xiāngjiāo / xiāngqiè / xiānglí |
| common chord | 公共弦 | gōnggòngxián |
Key points
- k = tan α = (y₂ − y₁)/(x₂ − x₁), 0° ≤ α < 180°; a vertical line has α = 90° and no slope.
- Forms: y − y₀ = k(x − x₀), y = kx + b, x/a + y/b = 1, Ax + By + C = 0 (k = −A/B); the intercept form misses lines through the origin.
- ∥: k₁ = k₂, b₁ ≠ b₂ (in general form A₁B₂ = A₂B₁ and not coincident); ⊥: k₁k₂ = −1 or A₁A₂ + B₁B₂ = 0.
- d = |Ax₀ + By₀ + C|/√(A² + B²); for parallel lines |C₁ − C₂|/√(A² + B²) with equal A and B.
- (x − a)² + (y − b)² = r²; x² + y² + Dx + Ey + F = 0 has centre (−D/2, −E/2) and radius √(D² + E² − 4F)/2.
- Line and circle: compare d with r; chord 2√(r² − d²); tangent x₀x + y₀y = r²; two circles: compare d with R + r and |R − r|.
Check yourself
12 questions. Every correct answer earns XP.
Topic test: 20 questions · 25 min
Finished the lesson? Check yourself with a timed test on this topic.