JEE Maths — Circles Complete Chapter Guide

Chapter Overview & Weightage

Circles is one of those chapters where JEE rewards consistency. Get the standard form and tangent conditions right, and you’ll pick up 2-3 marks every single paper. The chapter connects beautifully with coordinate geometry, so your investment here pays dividends in parabola, ellipse, and hyperbola too.

Circles typically carries 5-6% weightage in JEE Main — roughly 2 questions per paper. In JEE Advanced, expect 1 question, often combined with family of circles or locus problems. BITSAT allocates 3-4 questions directly.

Year	JEE Main (Questions)	JEE Advanced	Key Topics Asked
2024	2	1	Tangent from external point, radical axis
2023	2	1	Chord of contact, family of circles
2022	3	1	Orthogonal circles, normal at a point
2021	2	—	Common tangents, chord bisected at a point
2020	2	1	Equation of circle through three points
2019	2	1	Radical axis, power of a point

The pattern is clear: tangent conditions and family of circles are the highest-frequency subtopics. Radical axis appears almost every alternate year in Advanced.

Key Concepts You Must Know

Ranked by exam frequency — spend time in this order:

General equation $x^2 + y^2 + 2gx + 2fy + c = 0$ — centre $(-g, -f)$ , radius $\sqrt{g^2 + f^2 - c}$ . The condition $g^2 + f^2 - c > 0$ is tested directly in integer-type questions.
Tangent at a point on the circle — both the slope form and the “T = 0” form. Most students memorise T = 0 without understanding why it works; that understanding helps you adapt it for conics later.
Condition for tangency — for line $y = mx + c$ to touch $x^2 + y^2 = a^2$ , we need $c^2 = a^2(1 + m^2)$ . This is the most-tested single condition.
Chord of contact — equation of chord joining the two points of tangency from an external point. The formula $T = 0$ evaluated at the external point gives this directly.
Family of circles — $S_1 + \lambda S_2 = 0$ and $S + \lambda L = 0$ forms. These generate every circle through a fixed pair of points or through the intersection of a circle and a line.
Radical axis — locus of points having equal power with respect to two circles. Perpendicular to the line joining centres. The radical centre of three circles is a high-value JEE Advanced topic.
Orthogonal circles — condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ . Appears in 30-40% of JEE Advanced circle problems in some form.
Common tangents — number depends on relative position of circles. Memorise the cases by relative distance vs. sum/difference of radii.
Length of tangent from external point $(x_1, y_1)$ to circle $S = 0$ is $\sqrt{S_1}$ . Quick sanity check for MCQs.

Important Formulas

x^2 + y^2 + 2gx + 2fy + c = 0

Centre: $(-g, -f)$ , Radius: $r = \sqrt{g^2 + f^2 - c}$

For circle with centre $(h, k)$ and radius $r$ :

(x-h)^2 + (y-k)^2 = r^2

When to use: Any time the equation isn’t in standard form. Complete the square immediately — don’t work with the expanded form if you can avoid it.

Tangent at $(x_1, y_1)$ on $x^2 + y^2 = a^2$ :

xx_1 + yy_1 = a^2

Tangent at $(x_1, y_1)$ on general circle (T = 0 form):

xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0

Normal at any point passes through the centre — so the normal is just the line joining the centre to that point.

When to use: Tangent at a point uses T = 0. For tangent of given slope, use the condition of tangency and find $c$ .

From external point $(x_1, y_1)$ to circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$ :

xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0

This is literally T = 0 with $(x_1, y_1)$ substituted. Same formula, different interpretation.

When to use: When an external point is given and you need the chord joining the two tangent points.

Through intersection of circles $S_1 = 0$ and $S_2 = 0$ :

S_1 + \lambda S_2 = 0 \quad (\lambda \neq -1)

Through intersection of circle $S = 0$ and line $L = 0$ :

S + \lambda L = 0

Radical axis of $S_1$ and $S_2$ : $S_1 - S_2 = 0$

When to use: Whenever the problem says “a circle passing through the intersection of…” — write this family immediately, then apply the extra condition to find $\lambda$ .

Circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut orthogonally if:

2g_1g_2 + 2f_1f_2 = c_1 + c_2

When to use: Geometric meaning — the tangents at intersection points are perpendicular to each other, i.e., the radii to the point of intersection are perpendicular.

Length of tangent from $(x_1, y_1)$ to $S = 0$ :

L = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}

Power of point = $S_1$ (positive if outside, negative if inside, zero if on circle).

When to use: Whenever the problem asks for “length of tangent drawn from a point” — this is a one-line calculation.

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 January Shift 2

Problem: The equation of a circle which passes through the point $(1, -2)$ and $(4, -3)$ , and whose centre lies on the line $3x + 4y = 7$ , is:

Solution:

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ .

Since $(1, -2)$ lies on it:

1 + 4 + 2g - 4f + c = 0 \implies 2g - 4f + c = -5 \quad \cdots (1)

Since $(4, -3)$ lies on it:

16 + 9 + 8g - 6f + c = 0 \implies 8g - 6f + c = -25 \quad \cdots (2)

Subtract (1) from (2):

6g - 2f = -20 \implies 3g - f = -10 \quad \cdots (3)

Centre $(-g, -f)$ lies on $3x + 4y = 7$ :

-3g - 4f = 7 \quad \cdots (4)

From (3): $f = 3g + 10$ . Substitute in (4):

-3g - 4(3g + 10) = 7 \implies -3g - 12g - 40 = 7 \implies g = -3

Then $f = 3(-3) + 10 = 1$ .

From (1): $2(-3) - 4(1) + c = -5 \implies c = 5$ .

Circle: $x^2 + y^2 - 6x + 2y + 5 = 0$ , i.e., centre $(3, -1)$ , radius $\sqrt{9 + 1 - 5} = \sqrt{5}$ .

This is the standard “three conditions” setup. Always set up the general equation first, substitute each condition, and reduce to a system of equations. The centre-on-line condition is the third equation. This exact pattern appeared in 2022 as well.

PYQ 2 — JEE Main 2023 April Shift 1

Problem: If the tangents drawn from the point $(2, 3)$ to the circle $x^2 + y^2 = 5$ make angles $\alpha$ and $\beta$ with the $x$ -axis, find the equation of the chord of contact.

Solution:

The chord of contact from $(2, 3)$ to $x^2 + y^2 = 5$ is directly:

x \cdot 2 + y \cdot 3 = 5

2x + 3y = 5

That’s it. One line. The T = 0 formula does all the work here.

Let’s verify the point is external: $S_1 = 4 + 9 - 5 = 8 > 0$ . Yes, $(2,3)$ is outside the circle. Length of tangent = $\sqrt{8} = 2\sqrt{2}$ .

Many students try to find the actual tangent lines first, then find the chord joining the two contact points. That’s a 5-step detour. The chord of contact formula gives the answer in one step — always use T = 0 directly.

PYQ 3 — JEE Advanced 2022 Paper 2

Problem: The circles $x^2 + y^2 + 6x - 4y - 3 = 0$ and $x^2 + y^2 + 4x + 2y - 9 = 0$ intersect at points $A$ and $B$ . Find the equation of the circle with $AB$ as diameter.

Solution:

The line $AB$ (radical axis) is obtained by subtracting the two circle equations:

(x^2 + y^2 + 6x - 4y - 3) - (x^2 + y^2 + 4x + 2y - 9) = 0

2x - 6y + 6 = 0 \implies x - 3y + 3 = 0

Now, any circle through the intersection of $S_1 = 0$ and $S_2 = 0$ is:

S_1 + \lambda S_2 = 0

x^2 + y^2 + (6 + 4\lambda)x + (-4 + 2\lambda)y + (-3 - 9\lambda) = 0

For $AB$ to be a diameter, the centre must lie on $AB$ , i.e., on $x - 3y + 3 = 0$ :

Centre is $\left(-\frac{6+4\lambda}{2}, -\frac{-4+2\lambda}{2}\right) = \left(-(3+2\lambda), (2-\lambda)\right)$ .

Substituting in $x - 3y + 3 = 0$ :

-(3+2\lambda) - 3(2-\lambda) + 3 = 0

-3 - 2\lambda - 6 + 3\lambda + 3 = 0

\lambda = 6

Circle: $x^2 + y^2 + (6+24)x + (-4+12)y + (-3-54) = 0$

x^2 + y^2 + 30x + 8y - 57 = 0

The “circle with chord AB as diameter” trick: use family of circles $S_1 + \lambda S_2 = 0$ , then impose the condition that the centre lies on the radical axis $S_1 - S_2 = 0$ . This is a frequently tested technique in Advanced.

Difficulty Distribution

For JEE Main and Advanced, here’s how questions from this chapter typically split:

Difficulty	% of Questions	Typical Question Type
Easy	40%	Length of tangent, equation of tangent at a point, condition for tangency
Medium	45%	Chord of contact, family of circles, orthogonal circles with two circles given
Hard	15%	Radical centre, locus problems combining circles with other conics, advanced orthogonality

The easy questions are completely solvable in under 2 minutes if you know the formulas cold. That’s where you build your base score. Medium questions need the family of circles approach or chord of contact — these are the high-ROI questions to master.

Expert Strategy

Week 1 — Build the formula base. You need four formulas memorised with understanding: general equation conversions, T = 0 form (for tangent, chord of contact, and pair of tangents — same formula!), family of circles, and orthogonality condition. Do not try to derive them in the exam.

Week 2 — Pattern recognition. Circles problems are 80% pattern matching. The moment you see “circle passing through intersection of two circles/line,” write $S_1 + \lambda S_2 = 0$ or $S + \lambda L = 0$ . The moment you see “tangent from external point,” write T = 0. Train this reflex through PYQs.

The T = 0 formula is a Swiss Army knife. It gives you: (a) tangent at a point on the circle, (b) chord of contact from an external point, (c) the equation whose roots give the two contact points. One formula, three uses — understand this deeply and you’ll never confuse these three scenarios.

Connecting to other chapters: Circles + Straight Lines problems are very common — finding a chord bisected at a given point, or a chord of a given length. The approach: use $T = S_1$ (equation of chord bisected at $(x_1, y_1)$ is $T = S_1$ ). This template appears at least once every three years.

Time targets in the exam:

Easy circles question: 1.5 minutes
Medium circles question: 3 minutes
If you’re at 4 minutes on a circles problem, something’s wrong — reconsider your approach, don’t push deeper.

Common Traps

Trap 1 — Forgetting the radius condition. When finding the equation of a circle satisfying some geometric property, students often find the centre correctly but forget to verify that the given point actually lies on the circle (not just that the centre has the right coordinates). Write the radius equation explicitly.

Trap 2 — T = 0 for external vs. on-circle points. T = 0 gives the tangent at a point when the point is on the circle. If the point is external, T = 0 gives the chord of contact. These look identical algebraically but mean different things. Always check whether the given point satisfies the circle equation first.

Trap 3 — Sign error in orthogonality condition. The condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ , not $c_1 + c_2 = 0$ . The right side is $c_1 + c_2$ , not zero. Students memorise a partial version and lose marks. The geometric derivation: the Pythagorean theorem applied to the triangle formed by the two centres and either intersection point gives this condition.

Trap 4 — Radical axis direction. The radical axis is perpendicular to the line joining the centres, but it is NOT necessarily their perpendicular bisector (that would only be true for equal circles). When circles have different radii, the radical axis shifts towards the smaller circle.

Trap 5 — Family of circles $\lambda = -1$ case. The equation $S_1 + \lambda S_2 = 0$ represents all circles through the intersection of $S_1$ and $S_2$ except when $\lambda = -1$ , which gives the radical axis (a line, not a circle). If a problem asks you to find a circle in this family satisfying some property, and your $\lambda$ comes out as $-1$ , re-examine the problem — you may be looking for the radical axis itself, or the problem has a different setup.