The points (1, 1) and (3, −2) lie on the same side of the line 2x + y − 4 = 0 if:

Both substituted values have the same sign

Both substituted values are positive only

For two circles S₁ = 0 and S₂ = 0, the equation S₁ − S₂ = 0 represents:

The radical axis, a straight line perpendicular to the line of centres

The length of the tangent from an external point (x₁, y₁) to the circle S = 0 is:

√S₁, the square root of the value obtained by substituting the point

Straight Lines and Circles

Coordinate geometry is the part of the syllabus where a good choice of form does most of the work. A straight line has half a dozen equivalent equations, and picking the one that matches what you are given — two points, a point and a slope, two intercepts — turns a page of algebra into two lines. The same is true of circles, where the general equation x² + y² + 2gx + 2fy + c = 0 hides its centre and radius until you complete the square. The single most useful habit in this chapter is to write down the form that fits the data before starting, rather than forcing every problem through one remembered equation.

Forms of a Straight Line

Choose the form that matches the data
Given	Form to use
Slope and y-intercept	y = mx + c
A point and a slope	y − y₁ = m(x − x₁)
Two points	(y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
Both intercepts	x/a + y/b = 1
Perpendicular distance p from the origin	x cos α + y sin α = p
Distance from a point to a line	\|ax₁ + by₁ + c\|/√(a² + b²)

Two lines are parallel when their slopes are equal and perpendicular when m₁m₂ = −1. The angle between them satisfies tan θ = |(m₁ − m₂)/(1 + m₁m₂)|, which is undefined exactly when the lines are perpendicular — a consistency check rather than a problem. Three lines are concurrent when the determinant of their coefficients vanishes, which is much faster than solving two of them and substituting into the third.

🧠 The sign of ax + by + c tells you which side of the line a point is on

Substituting a point into ax + by + c gives a value whose sign is the same for every point on one side of the line and opposite on the other. So two points lie on the same side exactly when the two substituted values have the same sign, and on opposite sides when the signs differ — which is how questions about a line separating two points, or a point lying inside a triangle, are meant to be answered. Testing the three vertices of a triangle this way is far quicker than computing intersections, and the same idea extends to circles, where substituting into x² + y² + 2gx + 2fy + c gives a negative value inside, zero on, and positive outside.

Circles

The standard form (x − h)² + (y − k)² = r² shows the centre and radius directly. The general form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c), and it represents a real circle only when g² + f² − c > 0; if the expression is zero the circle degenerates to a point and if negative there is no real locus at all. Note that for a general second-degree equation to be a circle the coefficients of x² and y² must be equal and the xy term must be absent.

Position of a line relative to a circle
Condition on the distance d from the centre	Relationship
d > r	The line misses the circle
d = r	Tangent — this is the condition of tangency
d < r	The line cuts the circle in two points
Length of the tangent from an external point	√(x₁² + y₁² + 2gx₁ + 2fy₁ + c)
Two circles touch externally	Distance between centres = r₁ + r₂
Two circles touch internally	Distance between centres = \|r₁ − r₂\|

🎯 Why the condition of tangency is just a distance statement

A tangent touches a circle at exactly one point, and the shortest distance from the centre to any line is the perpendicular distance, so the line is tangent precisely when that perpendicular distance equals the radius. That single observation replaces the algebraic route of substituting the line into the circle and forcing the discriminant to zero — same answer, a fraction of the work. It also gives the tangent conditions for other curves once you know the perpendicular distance, and it explains why the tangent at a point on a circle is perpendicular to the radius drawn to that point, which is the fact most tangent questions actually turn on.

⚠️ Not every second-degree equation with x² and y² is a circle

Three conditions must all hold: the coefficients of x² and y² must be equal, the xy term must be absent, and g² + f² − c must be positive. Fail the first and you have an ellipse or hyperbola; fail the second and the axes are rotated; fail the third and there is no real curve. So a question giving 2x² + 2y² + 4x − 6y + 20 = 0 and asking for the radius is testing the third condition: dividing by 2 gives g = 1, f = −3/2, c = 10, and g² + f² − c is negative, so no such circle exists. The option list will contain a plausible radius obtained by taking the square root of the absolute value.

Two further constructions recur. The family of circles through the intersections of two circles S₁ = 0 and S₂ = 0 is S₁ + λS₂ = 0, and setting λ = −1 gives the radical axis, the common chord, whose equation is S₁ − S₂ = 0 and which is always perpendicular to the line of centres. And the equation of the tangent at a point (x₁, y₁) on a circle is obtained by T = 0, replacing x² by xx₁, y² by yy₁, x by (x + x₁)/2 and y by (y + y₁)/2 — a substitution rule that works for every conic, not just circles.

Key takeaways

Pick the line form that matches the data — two points, point and slope, or intercepts.
Lines are perpendicular when m₁m₂ = −1; three lines are concurrent when the coefficient determinant vanishes.
The sign of ax + by + c identifies which side of a line a point lies on.
x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c).
A real circle needs equal x² and y² coefficients, no xy term, and g² + f² − c > 0.
Tangency is simply the perpendicular distance from the centre equalling the radius.
Circles touch externally when the centre distance is r₁ + r₂ and internally when it is |r₁ − r₂|.
S₁ − S₂ = 0 is the radical axis and is always perpendicular to the line joining the centres.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The centre and radius of x² + y² − 4x + 6y − 12 = 0 are:
1. (2, −3) and 5
2. (−2, 3) and 5
3. (2, −3) and 12
4. (4, −6) and 5
Show answer
Answer: A — (2, −3) and 5
Comparing with x² + y² + 2gx + 2fy + c = 0 gives 2g = −4 so g = −2, and 2f = 6 so f = 3, hence the centre (−g, −f) = (2, −3). The radius is √(g² + f² − c) = √(4 + 9 + 12) = 5. The second option flips the signs by forgetting that the centre is the negative of (g, f), which is the most frequent slip in reading the general form.
A line is tangent to a circle when the perpendicular distance from the centre to the line is:
1. Less than the radius
2. Equal to the radius
3. Greater than the radius
4. Equal to the diameter
Show answer
Answer: B — Equal to the radius
A tangent meets the circle at exactly one point, and the closest approach of a line to the centre is its perpendicular distance, so equality with the radius is precisely the tangency condition. Less than the radius means the line cuts the circle twice and more than the radius means it misses entirely. Using the distance test avoids substituting the line into the circle and solving a quadratic discriminant.
x² + y² + 4x − 6y + 20 = 0 represents:
1. A circle of radius 3
2. A circle of radius √3
3. A single point
4. No real locus
Show answer
Answer: D — No real locus
Here g = 2, f = −3 and c = 20, so g² + f² − c = 4 + 9 − 20 = −7, which is negative and cannot be a squared radius, so no real points satisfy the equation. If the expression had been exactly zero the locus would be the single point (−2, 3). Checking the sign of g² + f² − c before quoting a radius is what separates the right answer from the plausible ones offered.
Two circles with radii 3 and 5 touch externally. The distance between their centres is:
1. 2
2. 8
3. 15
4. 4
Show answer
Answer: B — 8
External contact means the circles lie outside one another and meet at a single point on the line joining the centres, so that distance is the sum of the radii, 3 + 5 = 8. Internal contact would give the difference, |5 − 3| = 2, which is the first option — reading whether the contact is external or internal is therefore the whole question.
The points (1, 1) and (3, −2) lie on the same side of the line 2x + y − 4 = 0 if:
1. Both substituted values have the same sign
2. Both substituted values are positive only
3. Their sum is zero
4. The line passes between them
Show answer
Answer: A — Both substituted values have the same sign
Substituting a point into ax + by + c yields a value whose sign is constant on each side of the line, so two points share a side exactly when their values share a sign, whether both positive or both negative. Here (1,1) gives −1 and (3,−2) gives 0, so the second point actually lies on the line — a reminder to compute the values rather than assume. If the signs differ the line separates the points.
For two circles S₁ = 0 and S₂ = 0, the equation S₁ − S₂ = 0 represents:
1. Another circle
2. The radical axis, a straight line perpendicular to the line of centres
3. The line joining the centres
4. A pair of tangents
Show answer
Answer: B — The radical axis, a straight line perpendicular to the line of centres
Subtracting the two equations cancels the x² and y² terms, leaving a linear equation — the radical axis, which is the common chord when the circles intersect and is always perpendicular to the line joining the centres. It is a special case of the family S₁ + λS₂ = 0 with λ = −1, and it gives the common chord without ever finding the intersection points.
Three lines are concurrent when:
1. Their slopes are equal
2. The determinant of their coefficients is zero
3. Their intercepts are equal
4. They are mutually perpendicular
Show answer
Answer: B — The determinant of their coefficients is zero
Concurrency means all three pass through one common point, which is exactly the condition that the three linear equations have a common solution — and that happens when the coefficient determinant vanishes. Equal slopes would make them parallel, so they would never meet at all. Using the determinant is much faster than solving two equations and substituting into the third.
The length of the tangent from an external point (x₁, y₁) to the circle S = 0 is:
1. S₁ itself
2. √S₁, the square root of the value obtained by substituting the point
3. The distance to the centre
4. The radius
Show answer
Answer: B — √S₁, the square root of the value obtained by substituting the point
Because the tangent is perpendicular to the radius at the point of contact, Pythagoras gives (tangent length)² = (distance to centre)² − r², and expanding that difference is exactly the value of S at the point. So the length is √S₁, and the fact that S₁ is positive for an external point is the same criterion that tells you the point lies outside the circle in the first place.