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Circles Formulas for JEE 2027
The JEE Circles Formula Sheet covers the key formulas used to solve circle problems in coordinate geometry for JEE Main and JEE Advanced. It includes the standard and general equations of a circle, centre and radius, tangents, normals, chords, power of a point, radical axis, orthogonal circles and common tangents. Each topic includes formula tables, conditions for application and solved examples to explain how the results work.
Use this sheet to revise important concepts before practising JEE Circles Questions or attempting a JEE Mains mock test. The worked examples, shortcuts and common mistakes help you choose suitable formulas and check your calculations. For effective revision, understand the conditions behind each result, practise applying it independently and revisit formulas you find difficult to use.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$x,y$$ | Cartesian coordinates of a generic point | — |
| $$a,b$$ | Coordinates of a fixed point, usually centre or midpoint | — |
| $$h,k$$ | Centre of a circle in general form | — |
| $$r$$ | Radius of a circle | length |
| $$S(x,y)$$ | Quadratic expression $$x^{2}+y^{2}+2gx+2fy+c$$ | — |
| $$g,f$$ | Coefficients in general equation; centre $$(-g,-f)$$ | — |
| $$l,m$$ | Direction ratios of tangents/normals | — |
| $$d$$ | Distance (usually from a point to centre or to tangent) | length |
| $$S_{1}$$ | Value of $$S(x,y)$$ after substituting a specific point | — |
| $$T=0$$ | Line obtained by replacing $$x^{2},y^{2}$$ with $$xx_{1},yy_{1}$$ in $$S$$ | — |
| $$R_{a}$$ | Radical axis of two circles $$S_{1}=0$$ and $$S_{2}=0$$ | — |
| $$P$$ | Power of a point relative to circle $$S=0$$ | length2 |
| $$l_{t}$$ | Length of a tangent segment from a point | length |
| $$L$$ | Length of common tangent | length |
Equation of Circle – Standard, General, and Concentric Formulas
This sub-topic sets the language. Almost every later result plugs numbers into one of these forms. The examiner tests whether you can jump from the messy $$x^{2}+y^{2}+2gx+2fy+c=0$$ to the clean $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ without missing the sign. Build this muscle now; it saves you half a minute per problem.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ | Standard form centred at $$(h,k)$$ | $$r>0$$ |
| $$x^{2}+y^{2}+2gx+2fy+c=0$$ | General form $$S=0$$ | Discriminant $$g^{2}+f^{2}-c>0$$ |
| $$h=-g,\;k=-f,\;r=\sqrt{g^{2}+f^{2}-c}$$ | Centre and radius from general form | Same discriminant condition |
| Equation of circle through three points $$(x_{1},y_{1}),\,(x_{2},y_{2}),\,(x_{3},y_{3})$$: $$\begin{vmatrix}x&y;&x;^{2}+y^{2}&1\\x_{1}&y;_{1}&x;_{1}^{2}+y_{1}^{2}&1\\x_{2}&y;_{2}&x;_{2}^{2}+y_{2}^{2}&1\\x_{3}&y;_{3}&x;_{3}^{2}+y_{3}^{2}&1\end{vmatrix}=0$$ | Determinant form avoids solving simultaneous equations | Points non-collinear |
| Concentric circles: $$(x-h)^{2}+(y-k)^{2}=r_{1}^{2}$$(1), $$(x-h)^{2}+(y-k)^{2}=r_{2}^{2}$$(2) | Same centre, different $$r$$ | — |
| Difference of general forms of concentric pair gives $$c_{2}-c_{1}=r_{1}^{2}-r_{2}^{2}$$ | Jump directly to unknown $$c$$ when $$g,f$$ fixed | Only concentric family |
Worked Example
Find the centre and radius of the circle $$3x^{2}+3y^{2}-6x+12y-15=0$$, and state whether it is real.
- Divide by $$3$$ to reach standard coefficients: $$x^{2}+y^{2}-2x+4y-5=0$$.
- Compare with $$x^{2}+y^{2}+2gx+2fy+c=0$$. We get $$2g=-2\Rightarrow g=-1$$; $$2f=4\Rightarrow f=2$$; $$c=-5$$.
- Centre $$(-g,-f)=(1,-2)$$.
- Radius $$r=\sqrt{g^{2}+f^{2}-c}=\sqrt{(-1)^{2}+2^{2}-(-5)}=\sqrt{1+4+5}=\sqrt{10}$$.
- Discriminant positive, so the circle is real.
Answer: centre $$(1,-2)$$, radius $$\sqrt{10}$$, real circle.
Shortcuts and Special Cases
- If $$x$$ and $$y$$ appear only squared and multiplied by the same coefficient, complete the square mentally: $$x^{2}+y^{2}+ax+by+c=0$$ turns into $$(x+\tfrac{a}{2})^{2}+(y+\tfrac{b}{2})^{2}= \tfrac{a^{2}+b^{2}}{4}-c$$ without writing an extra line.
- A question quoting “circle with diameter AB” is simplest through midpoint and distance: centre $$=\bigl(\tfrac{x_{A}+x_{B}}{2},\tfrac{y_{A}+y_{B}}{2}\bigr)$$, radius $$=\tfrac{1}{2}\sqrt{(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}}$$.
- When $$c=0$$ in general form, the circle passes through the origin. That check often eliminates two options in MCQs.
Watch Out
- Forgetting the factor of 2: students write $$g=-2$$ instead of $$g=-1$$ when $$2g=-2$$.
- Using $$g^{2}+f^{2}+c$$ for radius instead of $$g^{2}+f^{2}-c$$.
- Trying to square-root a negative discriminant and inventing an imaginary radius—recognise that then there is no real circle.
- Determinant formula collapses if the three points are collinear; check by quick slope comparison first.
Position of a Point & Circle in Coordinate Geometry Formulas
Once you have the circle’s equation, almost every question asks “inside, on, or outside?”. This section gives the sign tests and distance comparisons that feed tangent conditions in the next topic. You already know $$S(x,y)$$; now you learn what its sign means.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$S_{1}=x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c$$ | Numeric value of general equation at $$(x_{1},y_{1})$$ | — |
| $$S_{1}<0,=0,>0$$ | Point inside, on, or outside the circle respectively | Real circle |
| Distance to centre $$d=\sqrt{(x_{1}-h)^{2}+(y_{1}-k)^{2}}$$ | Alternative test $$d<r$$ (inside) etc. | — |
| Shortest distance from a point outside the circle to its circumference: $$d-r$$ | Used in locus problems | $$d>r$$ |
| Length of tangent from point $$P$$: $$l_{t}=\sqrt{S_{1}}$$ | Directly usable in power of a point questions | $$S_{1}>0$$ |
Worked Example
Given the circle $$x^{2}+y^{2}-4x-6y+9=0$$, test whether the point $$P(2,1)$$ lies inside, on, or outside and find the length of the tangent from $$P$$.
- Compute $$S_{1}=2^{2}+1^{2}-4(2)-6(1)+9=4+1-8-6+9=0$$.
- Since $$S_{1}=0$$, the point lies on the circle.
- Length of tangent $$l_{t}=\sqrt{S_{1}}=0$$ — consistent with point on the circle.
Answer: P is on the circle; tangent length 0.
Shortcuts and Special Cases
- If sign testing time is short, compute $$g^{2}+f^{2}-c$$ once and store. Then $$S_{1}=(x_{1}+g)^{2}+(y_{1}+f)^{2}-r^{2}$$ avoids re-expansion.
- For symmetric points like $$(\pm a,0),(0,\pm a)$$, values come out in pairs—use symmetry to halve arithmetic.
Watch Out
- Mixing up inside/outside inequality when using $$d$$ method; always draw a tiny circle sketch next to your work.
- Plugging coordinates into $$S$$ after converting to standard form but still using $$g,f,c$$ of the original general form—values differ!
- For tangent length, forgetting to take square root and writing $$S_{1}$$ instead of $$\sqrt{S_{1}}$$.
Tangent to a Circle Formulas
The tangent line is the examiner’s favourite because it links slope, distance, and quadratic condition—three separate parts of the syllabus in one stroke. Every second set in a Mock Test slips in at least one of these formulas. Master them and most coordinate-geometry tangents fall in under a minute.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Point form: $$T=0$$ i.e. $$xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0$$ | Tangent at $$(x_{1},y_{1})$$ on circle $$S=0$$ | $$(x_{1},y_{1})$$ on circle |
| Slope form: $$y=mx\pm\sqrt{a^{2}m^{2}+a^{2}-r^{2}}$$ for circle $$x^{2}+y^{2}=r^{2}$$ | All tangents with slope $$m$$ | Inside √ must be non-negative |
| General line $$lx+my+n=0$$ touches circle $$S=0$$ when $$|l h+m k+n|=\sqrt{l^{2}+m^{2}}\;r$$ | Condition of tangency | Line not passing through centre |
| Angle between two tangents from point $$P$$: $$\theta$$ satisfies $$\sin\frac{\theta}{2}=\frac{r}{OP}$$ | Useful for min/max angle problems | $$OP>r$$ |
| Length of tangent from $$P$$ $$=\sqrt{OP^{2}-r^{2}}$$ | Reduces many quadratic eliminations | Outside point |
Worked Example
Find the equation of the tangent(s) to the circle $$x^{2}+y^{2}+4x-2y-4=0$$ having slope $$m=1$$.
- Convert to standard form: centre $$(-2,1)$$, radius $$r=\sqrt{4+1+4}=3$$.
- Line with slope 1: $$y= x + c$$.
- Distance from centre to this line equals radius:
$$\bigl|1\cdot(-2)-1+ c\bigr|/\sqrt{1^{2}+(-1)^{2}} =3$$ → $$|-3+c|/\sqrt{2}=3$$. - $$|-3+c|=3\sqrt{2}$$ → two values: $$c=3+3\sqrt{2}$$ or $$c=3-3\sqrt{2}$$.
- Equations: $$y=x+3+3\sqrt{2}$$ and $$y=x+3-3\sqrt{2}$$.
Answer: $$y=x+3\pm3\sqrt{2}$$.
Shortcuts and Special Cases
- Centre at origin: tangency condition reduces to $$|n|=r\sqrt{l^{2}+m^{2}}$$. One mental line, zero writing.
- If line given in intercept form $$\frac{x}{a}+\frac{y}{b}=1$$, touch condition is $$|ab|=r\sqrt{a^{2}+b^{2}}$$ — often spotted directly in MCQ options.
- For circle $$x^{2}+y^{2}=a^{2}$$, tangent at $$\theta$$ on circle is $$x\cos\theta+y\sin\theta=a$$; easy for parametric elimination.
Watch Out
- Forgetting absolute value in distance condition—sign flip loses one of the two tangents.
- Using $$h,k$$ of wrong circle after a quick copy-paste from earlier line in rough work.
- Skipping the division by $$\sqrt{l^{2}+m^{2}}$$ in distance formula; examiner loves inserting a line with exact same numerator but missing denominator.
Normal to a Circle Formulas
Normals appear less often than tangents but when they do, they separate the prepared student from the crowd. A normal is simply the line through the centre and the point of tangency. Knowing this unlocks fastest distance and locus solutions. This topic extends the tangency results you just mastered.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Normal at $$(x_{1},y_{1})$$ on $$S=0$$: line through centre $$C(h,k)$$ and point $$(x_{1},y_{1})$$ → $$y-y_{1}=\frac{y_{1}-k}{x_{1}-h}(x-x_{1})$$ | Equation of normal | $$(x_{1},y_{1})\neq(h,k)$$ |
| Slope of normal $$m_{n}=-\frac{1}{m_{t}}$$ | Perpendicular to tangent | Neither vertical line |
| Foot of perpendicular from centre to line $$Ax+By+C=0$$ is point of contact if line is tangent | Shortcut to find contact point directly | Tangency satisfied |
| Distance from centre to normal $$=0$$ by definition | Useful to locate normal in diagrams | — |
Worked Example
The tangent at point $$P$$ on circle $$x^{2}+y^{2}=25$$ has slope $$2$$. Find the equation of the normal at $$P$$.
- Tangent slope $$m_{t}=2$$ ⇒ normal slope $$m_{n}=-\tfrac{1}{2}$$.
- Parametric coordinates on circle: let $$P= (5\cos\theta,5\sin\theta)$$.
- Tangent equation by parametric form: $$5\cos\theta\,x+5\sin\theta\,y=25$$ ⇒ simplified to $$\cos\theta\,x+\sin\theta\,y=5$$.
- Slope of this tangent equals $$-\frac{\cos\theta}{\sin\theta}=-\cot\theta \stackrel{!}{=} 2$$ ⇒ $$\cot\theta=-2$$ ⇒ $$\tan\theta=-\tfrac{1}{2}$$.
- So $$\theta=\tan^{-1}(-\tfrac{1}{2})$$ (no need to compute value).
- Coordinates $$P=(5\cos\theta,5\sin\theta)$$.
- Normal passes through origin (centre) and $$P$$, slope $$-1/2$$ is consistent. Equation $$y=(-\tfrac{1}{2})x$$.
Answer: $$y=-\tfrac{1}{2}x$$.
Shortcuts and Special Cases
- For circles at origin, normal at $$(r\cos\theta,r\sin\theta)$$ is simply $$y\sin\theta - x\cos\theta=0$$ — one-line memory.
- Finding normal often avoids quadratic intersection entirely: draw centre-to-point vector, extend line.
Watch Out
- Confusing ‘normal’ with ‘normal chord’ (a chord perpendicular to radius) — read question carefully.
- For vertical tangent, normal slope is zero, not undefined.
Chords – Length, Chord of Contact, & Mid-point Formulas
Chords bridge tangency with intersection theory. Many JEE Circles Questions ask for locus of mid-point or length of chord cut by a given line. Master these three standard forms; they appear in direct one-step MSQ parts.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Length of chord with distance $$d$$ from centre: $$2\sqrt{r^{2}-d^{2}}$$ | Quick numeric chord length | $$d\lt r$$ |
| Chord of contact from external point $$P(x_{1},y_{1})$$: $$T=0$$ | Important for polar–pole problems | $$P$$ outside or on circle |
| Equation of chord with mid-point $$(x_{m},y_{m})$$: $$T=S_{1}$$ | Eliminates solving two-point form | — |
| Perpendicular bisector of chord passes through centre | Geometric property, used in constructions | — |
| Power of mid-point $$M$$: $$OM^{2}=r^{2}-\left(\frac{\text{chord length}}{2}\right)^{2}$$ | Reverse calculation | — |
Worked Example
A line $$3x-4y+5=0$$ cuts the circle $$x^{2}+y^{2}=25$$ at $$A$$ and $$B$$. Find length $$AB$$.
- Distance from centre to line: $$d=|3(0)-4(0)+5|/\sqrt{3^{2}+(-4)^{2}}=5/5=1$$.
- Chord length $$=2\sqrt{r^{2}-d^{2}}=2\sqrt{25-1}=2\sqrt{24}=4\sqrt{6}$$.
Answer: $$4\sqrt{6}$$.
Shortcuts and Special Cases
- If line is tangent, chord length collapses to zero—immediate check for whether quadratic will have equal roots.
- Mid-point form $$T=S_{1}$$ also gives chord bisected at a given point for any conic; same memory chunk works for parabola and ellipse.
- When diameter is asked, chord length formula with $$d=0$$ gives $$2r$$; a sanity check for bigger derivations.
Watch Out
- Squaring error: chord length formula demands $$r^{2}-d^{2}$$, not $$r^{2}+d^{2}$$.
- Using chord mid-point formula without first confirming mid-point is inside circle—question often embeds that trap.
Power of a Point, Radical Axis & Radical Centre Formulas
This trio ties two or three circles together elegantly. Examiner uses it to avoid solving simultaneous quadratics. Once you smell two circles in the same problem, check if radical axis will dispatch it quickly.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Power of $$P$$ w.r.t. $$S=0$$: $$P=S_{1}=OP^{2}-r^{2}$$ | Signed square of tangent length | — |
| Radical axis of $$S_{1}=0$$ and $$S_{2}=0$$: $$S_{1}-S_{2}=0$$ | Line of equal power | Circles need not intersect |
| Length of common chord = $$\frac{1}{d}\sqrt{(r_{1}^{2}+r_{2}^{2}-d^{2})^{2}-4r_{1}^{2}r_{2}^{2}}$$ | Optional, appears rarely | Real intersection $$|r_{1}-r_{2}|<d<r_{1}+r_{2}$$ |
| Radical centre: intersection of radical axes of any two pairs of three circles | Point of equal power to all three | Circles not coaxal |
| If $$S_{1}+kS_{2}=0$$, all such circles pass through points of intersection of $$S_{1}$$ and $$S_{2}$$ | Family of coaxal circles | — |
Worked Example
Find the radical axis of circles $$x^{2}+y^{2}-4x+2=0$$ and $$x^{2}+y^{2}+2x-3=0$$.
- Subtract equations: $$(x^{2}+y^{2}-4x+2)-(x^{2}+y^{2}+2x-3)=0$$.
- Simplify: $$-6x+5=0$$ → $$x=\frac{5}{6}$$.
Answer: $$x=\frac{5}{6}$$.
Shortcuts and Special Cases
- Equation difference instantly kills $$x^{2}+y^{2}$$ terms; do it mentally.
- If one circle is $$x^{2}+y^{2}=r^{2}$$, radical axis with another easily becomes linear in $$x,y$$ with no constant term—spot zero intercept in options.
- Any point where two tangents from external point $$P$$ touch two circles respectively lies on radical axis.
Watch Out
- Mixing up radical axis with common chord; they coincide only when circles intersect.
- Forgetting to write $$=0$$ after subtracting—line equation must be homogeneous.
System of Circles – Coaxal Family, Director Circle, Orthogonal Circles Formulas
This section upgrades power and radical ideas into families. The examiner asks you to find a circle passing through two fixed points and satisfying an extra condition. Recognise the “coaxal” keyword and many “JEE Circles Questions” shrink to plugging a value of $$k$$.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Coaxal family: $$S_{1}+kS_{2}=0$$ | All circles through intersection of $$S_{1}$$ and $$S_{2}$$ | $$k$$ scalar parameter |
| Director circle of $$x^{2}+y^{2}=r^{2}$$: $$x^{2}+y^{2}=2r^{2}$$ | Locus of points where two perpendicular tangents can be drawn | — |
| Condition for two circles orthogonal: $$2g g'+2f f'=c+c'$$ | Planar right-angle intersection | General forms $$S=0,S'=0$$ |
| Circle orthogonal to $$S=0$$ and passing through point $$P$$: $$S+ \lambda (x^{2}+y^{2}+2gx_{P}+2fy_{P}+c)=0$$ | Choose $$\lambda$$ to satisfy orthogonality | — |
Worked Example
Find the circle(s) orthogonal to $$x^{2}+y^{2}=25$$ and passing through $$A(4,3)$$.
- General circle through $$A$$: $$x^{2}+y^{2}+2gx+2fy+c=0$$ with constraint $$4^{2}+3^{2}+8g+6f+c=0$$ → $$25+8g+6f+c=0$$(i).
- Orthogonality with $$x^{2}+y^{2}=25$$: $$2g\cdot0+2f\cdot0=c+(-25)$$ ⇒ $$c=-25$$(ii).
- Plug (ii) into (i): $$25+8g+6f-25=0$$ → $$8g+6f=0$$ ⇒ $$4g+3f=0$$(iii).
- Choose $$g=3t,\;f=-4t$$. Then $$c=-25$$.
- Circle equation: $$x^{2}+y^{2}+6tx-8ty-25=0$$. Put $$A(4,3)$$ to find $$t$$: $$16+9+24t-24t-25=0$$ ⇒ 0=0 — any $$t$$ works. Infinite orthogonal circles through point.
Answer: $$x^{2}+y^{2}+6tx-8ty-25=0$$, $$t$$ any real.
Shortcuts and Special Cases
- If a circle is centred at origin, orthogonality with $$x^{2}+y^{2}=a^{2}$$ boils to radii relation $$r^{2}+a^{2}=d^{2}$$ where $$d$$ is distance between centres.
- Director circle for translated circle $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ has same centre and radius $$\sqrt{2}r$$ — quick memory tag.
Watch Out
- Forgetting both $$x^{2}+y^{2}$$ terms cancel in orthogonality condition derivation.
- Assuming a unique circle when family exists—always check remaining free parameter.
Common Tangents & Length of Common Tangent Formulas
External and internal tangents wrap up this topic by blending line condition with two circles simultaneously. MCQ setters love giving you four expressions for $$L$$; pick the correct one without re-deriving. Practise these to shave off time.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Distance between centres $$d=\sqrt{(h-h')^{2}+(k-k')^{2}}$$ | Base quantity | — |
| Number of common tangents: $$\begin{cases}4,& d>r_{1}+r_{2}\\3,& d=r_{1}+r_{2}\\2,& |r_{1}-r_{2}|<d<r_{1}+r_{2}\\1,& d=|r_{1}-r_{2}|\\0,& d<|r_{1}-r_{2}|\end{cases}$$ | Quick classification | $$r_{1}\ge r_{2}$$ assumed |
| Length of direct common tangent: $$L=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$ | When external tangents exist | $$d>r_{1}-r_{2}$$ |
| Length of transverse common tangent: $$L=\sqrt{d^{2}-(r_{1}+r_{2})^{2}}$$ | Internal tangents | $$d>r_{1}+r_{2}$$ |
| Equation of direct common tangent when centres on $$x$$-axis: $$y=\pm \frac{r_{1}-r_{2}}{\sqrt{d^{2}-(r_{1}-r_{2})^{2}}}(x-x_{C})+y_{C}$$ | Derived slope form | Centres aligned horizontally |
Worked Example
Two circles $$x^{2}+y^{2}=16$$ and $$(x-10)^{2}+y^{2}=9$$. Find length of their direct common tangent.
- Centres $$C_{1}(0,0),C_{2}(10,0)$$, so $$d=10$$.
- Radii $$r_{1}=4,r_{2}=3$$.
- $$L=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}=\sqrt{100-(1)^{2}}=\sqrt{99}$$.
Answer: $$\sqrt{99}$$.
Shortcuts and Special Cases
- When both circles equal radii, direct and transverse tangent formulas merge: $$L=\sqrt{d^{2}-4r^{2}}$$.
- If one circle lies inside another (concentric or not) with no tangent, answer for common tangents is instantly zero—no further work.
Watch Out
- Using $$r_{1}+r_{2}$$ instead of $$r_{1}-r_{2}$$ in direct tangent formula.
- Forgetting square root; examiner offers $$d^{2}-(r_{1}-r_{2})^{2}$$ as distractor.
- Centre distance mis-computed when circles have negative coordinates—square and sum, no sign worry.
Commonly Confused JEE Circles Formulas and Their Differences
| Formula 1 | Formula 2 | Tell them apart by |
|---|---|---|
| Tangent length $$\sqrt{S_{1}}$$ | Chord half-length $$\sqrt{r^{2}-d^{2}}$$ | First uses external point, second uses line-circle distance. |
| Radical axis $$S_{1}-S_{2}=0$$ | Tangent at point $$T=0$$ | Radical axis contains no $$x^{2},y^{2}$$ terms from start; tangent got them canceled after substitution. |
| Direct tangent length $$\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$ | Transverse tangent length $$\sqrt{d^{2}-(r_{1}+r_{2})^{2}}$$ | Look at sign between radii: minus for direct, plus for transverse. |
| Orthogonality $$2g g'+2f f'=c+c'$$ | Equal radii concentric difference $$c_{2}-c_{1}=r_{1}^{2}-r_{2}^{2}$$ | Orthogonality involves both sets of $$g,f$$; concentric difference does not. |
Common Mistakes to Avoid When Solving JEE Circles Formulas
- Dropping the factor 2 in $$2g,2f$$ while shifting between general and standard forms.
- Evaluating $$S_{1}$$ with rounded centre coordinates instead of exact fractions, creating wrong sign classification.
- Applying direct tangent formula when circles overlap; under-root becomes negative and student still writes real answer.
- Forgetting to check whether three points are collinear before using determinant circle through three points.
- Mixing radical axis with common chord and hence assuming intersection points exist.
- Assuming orthogonality implies equal radii—it does not.
- Skipping absolute value while using distance formula for tangency condition.
JEE Circles Formulas for Quick Revision
- $$(x-h)^{2}+(y-k)^{2}=r^{2}$$ — standard circle, centre $$(h,k)$$, radius $$r$$.
- Radius from general: $$r=\sqrt{g^{2}+f^{2}-c}$$.
- Point test $$S_{1}\lt0,=0,\gt0$$ → inside, on, outside.
- Tangent at $$(x_{1},y_{1})$$: $$T=0$$.
- Line $$lx+my+n=0$$ tangent ⇔ $$|lh+mk+n|=r\sqrt{l^{2}+m^{2}}$$.
- Chord length with centre distance $$d$$: $$2\sqrt{r^{2}-d^{2}}$$.
- Power of point: $$S_{1}=OP^{2}-r^{2}$$, tangent length $$\sqrt{S_{1}}$$.
- Radical axis of $$S_{1},S_{2}$$: $$S_{1}-S_{2}=0$$.
- Orthogonality: $$2g g'+2f f'=c+c'$$.
- Direct common tangent length: $$\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$; transverse: $$\sqrt{d^{2}-(r_{1}+r_{2})^{2}}$$.
- Director circle: $$x^{2}+y^{2}=2r^{2}$$ for centre at origin.
- Coaxal family through two points: $$S_{1}+kS_{2}=0$$.
- Mid-point chord: $$T=S_{1}$$.
- Angle between tangents from $$P$$: $$\sin\frac{\theta}{2}=\frac{r}{OP}$$.
- This sheet + practice = full mark section in your “JEE Study Material”.
Circles Formulas for JEE 2027
Revising circle equations, tangents, normals, chords and radical axes builds a strong foundation for coordinate geometry problems. Use this JEE Circles Formula Sheet to review key results and understand the conditions for applying them. Follow each topic with practice questions to identify mistakes and improve your accuracy.
As you revise JEE Mains Maths Important Chapters, connect circles with straight lines and other coordinate geometry topics through mixed practice. Keep a JEE Maths Formula Sheet handy to revisit difficult formulas and record common errors. Regular revision, previous-year questions and timed practice will help you apply these concepts more confidently.
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