1. Chapter at a Glance
A circle is the set of all points in a plane at a fixed distance (radius) from a fixed point (centre). In coordinate geometry a circle is a second-degree equation in which x² and y² have equal coefficients and there is no xy term. JEE Main usually asks one question on circles — on tangents, chords, two circles or a locus.
- Centre and radius from the general equation; circles through given points.
- Condition of tangency; length of tangent; chord of contact.
- Position of two circles and the number of common tangents.
- Orthogonal circles and length of a chord.
2. Equations of a Circle
| Form | Equation |
|---|---|
| Centre (h, k), radius r | (x − h)² + (y − k)² = r² |
| General form | x² + y² + 2gx + 2fy + c = 0, centre (−g, −f), r = √(g² + f² − c) |
| Diameter form (ends A, B) | (x − x1)(x − x2) + (y − y1)(y − y2) = 0 |
| Parametric form | x = h + r cos θ, y = k + r sin θ |
The general equation represents a real circle if g² + f² − c > 0, a point circle if it is 0, and no real circle if it is negative.
Special positions
- Touching the x-axis: radius = |f| (so g² = c). Touching the y-axis: radius = |g| (so f² = c).
- Through the origin: c = 0.
- Circle through three points: substitute them in the general form and solve for g, f, c.
3. Tangents, Normals and Chords
We use the short notation S = x² + y² + 2gx + 2fy + c, S1 = x1² + y1² + 2gx1 + 2fy1 + c, and T = xx1 + yy1 + g(x + x1) + f(y + y1) + c.
- Position of a point: S1 > 0 outside, = 0 on, < 0 inside the circle.
- Tangent at (x1, y1) on the circle: T = 0. For x² + y² = a²: xx1 + yy1 = a².
- Tangent with slope m to x² + y² = a²: y = mx ± a√(1 + m²).
- Condition of tangency: y = mx + c touches x² + y² = a² if c² = a²(1 + m²). In general, a line is a tangent if its distance from the centre equals the radius.
- Length of tangent from an outside point: √S1.
- Chord of contact from an outside point, and chord with given midpoint (x1, y1): T = 0 and T = S1 respectively.
- Every normal passes through the centre.
Length of a chord
If a chord is at distance d from the centre of a circle of radius r, its length is 2√(r² − d²).
4. Two Circles
Let the centres be C1, C2, radii r1, r2, and d = C1C2.
| Condition | Position | Common tangents |
|---|---|---|
| d > r1 + r2 | Separate (outside each other) | 4 |
| d = r1 + r2 | Touch externally | 3 |
| |r1 − r2| < d < r1 + r2 | Intersect at two points | 2 |
| d = |r1 − r2| | Touch internally | 1 |
| d < |r1 − r2| | One inside the other | 0 |
- Common chord / radical axis: S1 − S2 = 0 (with both equations in standard form).
- Orthogonal circles (tangents at an intersection point are perpendicular): 2g1g2 + 2f1f2 = c1 + c2, equivalently d² = r1² + r2².
- Family of circles through the intersection of a circle S = 0 and a line L = 0: S + λL = 0; through two circles: S1 + λS2 = 0 (λ ≠ −1).
5. Formula Master Table
| Result | Formula |
|---|---|
| Centre and radius | (−g, −f), √(g² + f² − c) |
| Tangent at a point | T = 0 |
| Tangency condition | c² = a²(1 + m²) |
| Length of tangent | √S1 |
| Chord of contact | T = 0 |
| Chord with midpoint (x1, y1) | T = S1 |
| Chord length | 2√(r² − d²) |
| Radical axis / common chord | S1 − S2 = 0 |
| Orthogonal circles | 2g1g2 + 2f1f2 = c1 + c2 |
| Pair of tangents from (x1, y1) | SS1 = T² |
6. Mind Map
- Centre–radius
- General form
- Diameter form
- S1 sign → position
- Tangency: distance = r
- Chord length
- T = 0
- √S1
- T = S1
- d vs r1 ± r2
- Common tangents
- Orthogonal, radical axis
7. Tips and Tricks
- To test tangency of any line, compare its distance from the centre with the radius — faster than solving simultaneously.
- If the equation is given as ax² + ay² + …, divide by a before reading g, f and c.
- A circle through the origin cutting intercepts a and b on the axes: x² + y² − ax − by = 0.
- For the circle with the segment joining (x1, y1) and (x2, y2) as diameter, use the diameter form directly — no need to find the centre.
- Reading the centre as (g, f) instead of (−g, −f).
- Using S1 − S2 = 0 when the coefficients of x² in the two equations are not both 1.
- Counting common tangents for intersecting circles as 4 — it is 2.
8. Important One-liners
- The angle in a semicircle is a right angle.
- The perpendicular from the centre to a chord bisects the chord.
- Tangents drawn from an external point to a circle are equal in length.
- The radius is perpendicular to the tangent at the point of contact.
- The locus of the midpoints of chords of x² + y² = a² that subtend a right angle at the centre is x² + y² = a²/2.
- The director circle of x² + y² = a² (locus of points from which the tangents are perpendicular) is x² + y² = 2a².
- If two circles touch, the point of contact lies on the line joining their centres.
9. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
Find the radius of the circle x² + y² − 4x + 6y − 12 = 0.
Show solution
g = −2, f = 3, c = −12. Centre (2, −3); r = √(4 + 9 + 12) = √25 = 5.
Answer: 5
Find the length of the tangent from the point (4, 5) to the circle x² + y² = 16.
Show solution
Length = √S1 = √(16 + 25 − 16) = √25 = 5.
Answer: 5
The circle passing through (0, 0), (4, 0) and (0, 6) has radius:
- √13
- 5
- √26
- √52
Show solution
Through the origin with intercepts 4 and 6: x² + y² − 4x − 6y = 0. Centre (2, 3); r = √(4 + 9) = √13.
(Also: the chord from (4, 0) to (0, 6) is a diameter, because the axes meet at a right angle at the origin; its length is √52 = 2√13.)
Answer: (A)
Find the number of common tangents to the circles x² + y² = 9 and x² + y² − 8x + 7 = 0.
Show solution
First circle: centre (0, 0), r1 = 3. Second: centre (4, 0), r2 = √(16 − 7) = 3.
d = 4, r1 + r2 = 6, |r1 − r2| = 0. Since 0 < 4 < 6 the circles intersect at two points, so there are 2 common tangents.
Answer: 2
The length of the chord cut off by the line x + y = 1 from the circle x² + y² = 4 is:
- √14
- √7
- 2√2
- 4
Show solution
Distance of the centre (0, 0) from x + y − 1 = 0 is d = 1/√2, so d² = 1/2.
Chord = 2√(r² − d²) = 2√(4 − 1/2) = 2√(7/2) = √14.
Answer: (A)
The line y = mx + c touches the circle x² + y² = a² if:
- c = a(1 + m²)
- c² = a²(1 + m²)
- c² = a²(1 − m²)
- c = am
Show solution
The distance from the centre (0, 0) to mx − y + c = 0 must equal a: |c|/√(1 + m²) = a ⇒ c² = a²(1 + m²).
Answer: (B)
If the circles x² + y² − 4x + 2y + k = 0 and x² + y² − 6y − 3 = 0 cut orthogonally, find k.
Show solution
First circle: g1 = −2, f1 = 1, c1 = k. Second: g2 = 0, f2 = −3, c2 = −3.
2g1g2 + 2f1f2 = c1 + c2 ⇒ 0 + 2(1)(−3) = k − 3 ⇒ k = −3.
Check: the first circle then has radius √(4 + 1 + 3) = √8, a real circle.
Answer: −3
The locus of the midpoints of chords of x² + y² = 16 that subtend a right angle at the centre is:
- x² + y² = 8
- x² + y² = 4
- x² + y² = 32
- x² + y² = 12
Show solution
A chord subtending 90° at the centre forms an isosceles right triangle with two radii of length 4. The distance of the chord from the centre is 4 cos 45° = 2√2.
The midpoint of the chord is at this distance from the centre, so its locus is x² + y² = (2√2)² = 8.
Answer: (A)
10. Quick Revision Summary
- Centre (−g, −f), radius √(g² + f² − c).
- Tangent at a point: T = 0; tangency: distance from centre = radius.
- Tangent length √S1; chord with midpoint: T = S1; chord length 2√(r² − d²).
- Two circles: compare d with r1 + r2 and |r1 − r2| → 4, 3, 2, 1, 0 common tangents.
- Orthogonal: 2g1g2 + 2f1f2 = c1 + c2.