USE Parabola whose axis is vertical (parallel to Y-axis).
🎯
Position of a Point w.r.t. Parabola
3 conditions▼
S₁ Test for $y^2-4ax=0$
$S_1 = y_1^2 - 4ax_1$
$S_1 < 0 \Rightarrow$ Point INSIDE the parabola (0 tangents can be drawn)
$S_1 = 0 \Rightarrow$ Point ON the parabola (1 tangent)
$S_1 > 0 \Rightarrow$ Point OUTSIDE (2 tangents can be drawn)
USE Instantly classify a point's position without graph — prerequisite for pair of tangents.
Key: Make coefficient of $x^2$ and $y^2$ non-negative before testing.
📏
Focal Distance / Focal Radii
4 formulas▼
Horizontal Parabolas
Focal distance for $y^2 = \pm4ax$
$$\text{Focal distance of } P(h,k) = |h| + a$$
USE Distance from point on parabola to focus = distance to directrix = $|h|+a$.
Vertical Parabolas
Focal distance for $x^2 = \pm4ay$
$$\text{Focal distance of } P(h,k) = |k| + a$$
USE Same idea for vertically oriented parabolas.
Shifted Parabola
$(y-\beta)^2 = \pm4a(x-\alpha)$
$$\text{Focal distance} = |h-\alpha| + a, \quad a>0$$
USE Focal distance for translated parabola — shift then apply standard formula.
Min focal distance on any parabola = $a$ (at vertex). If focal distance < $a$: impossible; = $a$: vertex only; > $a$: two such points.
🔢
Parametric Coordinates
6 forms▼
All Standard Parametric Forms
$y^2=4ax$: point $\equiv (at^2,\; 2at)$
$y^2=-4ax$: point $\equiv (-at^2,\; 2at)$
$x^2=4ay$: point $\equiv (2at,\; at^2)$
$x^2=-4ay$: point $\equiv (2at,\; -at^2)$
$(y-\beta)^2=4a(x-\alpha)$: point $\equiv (\alpha+at^2,\; \beta+2at)$
$(x-\alpha)^2=4a(y-\beta)$: point $\equiv (\alpha+2at,\; \beta+at^2)$
USE Reduce chord/tangent/normal problems to single variable $t$ — massive simplification in JEE Advanced.
USE Find common tangent by combining parabola tangency condition with circle's perpendicular-radius rule.
Properties of tangents: Foot of ⊥ from focus to any tangent lies on tangent at vertex. Image of focus in any tangent lies on directrix. Tangent between contact point & directrix subtends right angle at focus.
USE HM of segments of any focal chord = semi-latus rectum.
Reflection Law (Optical Property)
$$\text{Ray from one focus reflects off ellipse and passes through other focus}$$
$$\frac{PS'}{PS} = \frac{S'N}{NS} \quad \text{(angle bisector property of normal)}$$
USE Ellipse is a perfect mirror — used in whispering gallery, lithotripsy problems.
Image of Focus in Tangent
$$\text{Image of one focus in any tangent lies on the line joining other focus and point of contact}$$
USE Find tangent point when focus and image of focus are given.
Locus of Tangent-Directrix Intersection
$$\text{Tangent between contact point and directrix subtends right angle at corresponding focus}$$
USE Property V — prove specific angle results in JEE Advanced.
⟆ Hyperbola
eccentricity e > 1 | |PS − PS'| = 2a | h² > ab
Core Formula
General / Definition
Note / Trick
📋
Standard Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
USE R lies outside segment — if λ in assumed ratio λ:1 is negative, division is external.
Harmonic Conjugate
$$\frac{2}{AB} = \frac{1}{AP} + \frac{1}{AQ} \quad \text{(AP, AB, AQ in H.P.)}$$
USE P divides AB internally and Q divides AB externally in same ratio m:n → P and Q are harmonic conjugates.
Centroid (G)
Centroid Formula
$$G = \left(\frac{x_1+x_2+x_3}{3},\;\frac{y_1+y_2+y_3}{3}\right)$$
Centroid divides each median in ratio $2:1$ from vertex.
USE G is always inside the triangle — use to find third vertex when G and two vertices are given.
Incentre (I) — where angle bisectors meet
Incentre Formula
$$I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c},\;\frac{ay_1+by_2+cy_3}{a+b+c}\right)$$
where $a=BC,\; b=CA,\; c=AB$ (sides opposite to vertices $A,B,C$)
USE Centre of inscribed circle (incircle). Angle bisector from A divides BC in ratio $c:b$ (ratio of adjacent sides).
Circumcentre (O) — equidistant from all vertices
Circumcentre — Key Properties
$OA^2 = OB^2 = OC^2 = R^2$
$$O = \left(\frac{x_1\sin 2A+x_2\sin 2B+x_3\sin 2C}{\sin 2A+\sin 2B+\sin 2C},\;\frac{y_1\sin 2A+y_2\sin 2B+y_3\sin 2C}{\sin 2A+\sin 2B+\sin 2C}\right)$$
Short-cut: If right-angled at B, circumcentre = midpoint of hypotenuse AC.
USE Fastest method: set $OA^2=OB^2$ and $OB^2=OC^2$, solve two equations for $(h,k)$.
Orthocentre (H) — altitudes meet
Orthocentre — Key Properties
$$H = \left(\frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C},\;\frac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A+\tan B+\tan C}\right)$$
Right-angled: H is at the vertex with right angle. Equilateral: G = I = O = H (all four coincide). Euler Line: O, G, H are collinear; $G$ divides $OH$ in ratio $1:2$.
USE Euler line: once circumcentre O and centroid G known, orthocentre $H = 3G - 2O$ (vector form).
USE Split any polygon into triangles and sum areas, or directly apply the shoelace formula going around the vertices in order.
KEY FACT: Area = 0 → points are collinear. Vertices of equilateral triangle cannot all have rational coordinates (area would be $\frac{\sqrt{3}}{4}a^2$ = irrational, but rational × rational combinations = rational — contradiction).
〰
Locus — Finding the Equation
3 formulas▼
Standard Procedure
Step 1: Let moving point $P = (h,k)$
Step 2: Express the geometric condition in terms of $h,k$ and any parameters
Step 3: Eliminate all parameters
Step 4: Replace $h \to x$, $k \to y$ — the result is the locus equation
USE General method — the key step is elimination of the free parameter from the geometric condition.
Rod on Axes Locus (classic)
Rod of length $\ell$ with ends on axes, point dividing in $m_1:m_2$:
$$\left(\frac{m_1 x}{m_2}\right)^2 + \left(\frac{m_2 y}{m_1}\right)^2 = \ell^2 \quad\Longrightarrow\quad m_1^2x^2+m_2^2y^2 = \left(\frac{m_1m_2\ell}{m_1+m_2}\right)^2$$
USE Classic locus — point on rod with ends on axes traces an ellipse (or circle if divides equally).
cotA + cotB = λ Locus
Fixed points $A(a,0)$, $B(-a,0)$; vertex $C(h,k)$ with $\cot A + \cot B = \lambda$:
$$\cot A = \frac{a-h}{k},\quad \cot B = \frac{a+h}{k} \Rightarrow \frac{2a}{k}=\lambda$$
Locus: $\boxed{y\lambda = 2a}$ (a horizontal line)
USE Whenever cotA + cotB = constant, vertex traces a horizontal line parallel to AB.
Convert to normal form: divide by $\sqrt{a^2+b^2}$ (after making RHS positive)
USE Every first-degree equation in $x,y$ represents a straight line — proven because any point on the segment joining two solutions also satisfies it.
Parametric Form
$$\frac{x-x_1}{\cos\theta} = \frac{y-y_1}{\sin\theta} = r \;\Longrightarrow\; P=(x_1+r\cos\theta,\;y_1+r\sin\theta)$$
$|r|$ = distance of $P$ from fixed point $(x_1,y_1)$; $r>0$ forward, $r<0$ backward along direction $\theta$.
USE Find intersection distances: substitute parametric coords into the other curve, get equation in $r$ — product of roots $= r_1 r_2$ = PQ · PR.
Parallel & Perpendicular Lines
From a Given Line $ax+by+c=0$
Parallel: $ax + by + \lambda = 0$ (same coefficients, different constant)
Perpendicular: $bx - ay + k = 0$ (swap coefficients, flip sign of one)
USE $\lambda$ or $k$ determined from additional condition (passes through a point, distance = $d$, etc.).
∠
Angle, Distance & Position
8 formulas▼
Angle Between Two Lines
Angle Formula
$$\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| \quad \text{(acute angle)}$$
Internal angles of triangle with sides $m_1>m_2>m_3$:
$$\tan A = \frac{m_1-m_2}{1+m_1m_2},\quad \tan B=\frac{m_2-m_3}{1+m_2m_3},\quad \tan C=\frac{m_3-m_1}{1+m_3m_1}$$
Negative $\tan C$ → obtuse triangle.
USE Take slopes in decreasing order $m_1>m_2>m_3$; third tangent value automatically has opposite sign if obtuse.
Line at Angle $\alpha$ to Given Line
$$y - y_1 = \frac{m\pm\tan\alpha}{1\mp m\tan\alpha}(x-x_1) \quad \text{(two lines making angle }\alpha\text{ with slope }m)$$
USE When line through a given point makes a specific angle with another line — gives two solutions (±).
Position of Points
Same / Opposite Side of a Line
For line $ax+by+c=0$ and points $P(x_1,y_1)$, $Q(x_2,y_2)$:
Same side if $(ax_1+by_1+c)$ and $(ax_2+by_2+c)$ have same sign.
Opposite side if signs differ.
USE Determine which region a point lies in — used for triangle inside/outside checks and inequality solution regions.
Perpendicular Distance
Point to Line Distance
$$d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$$
From origin: $\displaystyle P = \frac{|c|}{\sqrt{a^2+b^2}}$
USE Core formula — radius of inscribed circle equals distance from incentre to any side.
Distance Between Parallel Lines
$$d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} \quad \text{(for }ax+by+c_1=0 \text{ and } ax+by+c_2=0\text{)}$$
Area of parallelogram bounded by $y=m_1x+c_1$, $y=m_1x+c_2$, $y=m_2x+d_1$, $y=m_2x+d_2$:
$$\text{Area} = \left|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}\right|$$
USE Coefficient of $x$ and $y$ must be identical in both lines before applying — normalise first if needed.
Foot of Perpendicular & Reflection
Foot of Perpendicular from $(x_1,y_1)$ to $ax+by+c=0$
$$\frac{h-x_1}{a} = \frac{k-y_1}{b} = \frac{-2(ax_1+by_1+c)}{a^2+b^2}$$
Special reflections of $P(x,y)$: x-axis→$(x,-y)$; y-axis→$(-x,y)$; origin→$(-x,-y)$; $y=x$→$(y,x)$
USE Image = $2\times\text{foot} - \text{original point}$. The factor $-2$ vs $-1$ in foot formula is the only difference.
Algebraic Sum of Perpendiculars = 0
$$\sum_{i=1}^{n}\frac{ax_i+by_i+c}{\sqrt{a^2+b^2}} = 0 \;\Rightarrow\; \text{variable line passes through fixed point }\left(\frac{\sum x_i}{n},\frac{\sum y_i}{n}\right)$$
USE If algebraic sum of perpendiculars from n given points to a variable line is zero, the line always passes through the centroid of those n points.
⊕
Family of Lines, Concurrency & Determinant Form
6 formulas▼
Family of Lines Through Intersection of $L_1=0$ and $L_2=0$
$$L_1 + \lambda L_2 = 0 \quad \text{(for all values of }\lambda\text{)}$$
i.e. $a_1x+b_1y+c_1 + \lambda(a_2x+b_2y+c_2) = 0$
USE Any line through intersection point of two given lines — find $\lambda$ from extra condition (passes through a point, perpendicular to one, etc.).
Fixed Point of a Family (ax+by+c=0 with constraint)
If $pa+qb+rc=0$, rearrange as $L_1+\frac{r}{q}L_2=0$ form → fixed point = intersection of $L_1=0$, $L_2=0$.
Quick method: treat as $\lambda L_1 + \mu L_2 = 0$ and find their intersection.
USE E.g. $3a+2b+5c=0$: rewrite $ax+by+c=0$ eliminating one coefficient → passes through $(\frac{3}{5},\frac{2}{5})$.
Concurrency of Three Lines
$$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$$
Equivalently: $\exists\; l,m,n$ (not all zero) such that $lL_1+mL_2+nL_3=0$.
USE Determinant = 0 is necessary but not sufficient (may be parallel) — verify by checking the system is consistent.
USE Clean form for median, altitude through a triangle vertex using midpoint coordinates.
Family Parameterised by Angle (θ parameter)
If family is $(A\sin\theta+B\cos\theta)x+(C\sin\theta+D\cos\theta)y+(E\sin\theta+F\cos\theta)=0$,
rewrite as $\sin\theta(\cdot)+\cos\theta(\cdot)=0 \;\Rightarrow\; L_1+\tan\theta\cdot L_2=0$
→ passes through intersection of $L_1=0$ and $L_2=0$.
USE Separate sine and cosine terms; treat as $P+\lambda Q=0$ family — fixed point is intersection of P=0, Q=0.
⊿
Angle Bisectors of Two Lines
6 formulas▼
Bisector Equation
$$\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$
Two bisectors — one acute, one obtuse (they are always perpendicular to each other).
USE Two bisectors always intersect at right angles — use this to find the second once the first is known.
Bisector Containing the Origin
Make $c_1>0$ and $c_2>0$ (multiply equation by $-1$ if constant term is negative).
Then the bisector containing origin is the positive sign bisector.
USE Origin lies in the acute-angle region if $a_1a_2+b_1b_2 < 0$ (after making constants positive).
USE Alternative: find angle between a bisector and one given line; $\tan\phi<1\Rightarrow\phi<45°\Rightarrow$ that bisector is the acute one.
Bisector Containing a Given Point $(x_1,y_1)$
Compute $L_1(x_1,y_1)=a_1x_1+b_1y_1+c_1$ and $L_2(x_1,y_1)$.
If both have same sign → use $+$ bisector.
If signs differ → use $-$ bisector.
USE Directly test which bisector the given point satisfies — no need to compute angles.
⚡
Pair of Straight Lines (Homogeneous & General)
15 formulas▼
Homogeneous Second Degree: $ax^2+2hxy+by^2=0$
Sum & Product of Slopes
$$m_1+m_2 = -\frac{2h}{b},\quad m_1m_2 = \frac{a}{b}$$
$$\tan\theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right| \quad \text{(angle between the pair)}$$
USE Both lines always pass through the origin. Treat as quadratic in $m = y/x$.
Nature of Lines
$h^2 - ab > 0$: real and distinct
$h^2 - ab = 0$: real and coincident
$h^2 - ab < 0$: imaginary (complex conjugates, only real intersection = origin)
USE Same discriminant logic as quadratic — $h^2-ab$ replaces $b^2-4ac$.
Special Conditions
Perpendicular: $a+b=0$ (sum of coefficients of $x^2$ and $y^2$ = 0)
Coincident: $h^2=ab$
Equally inclined to x-axis: $h=0$ (no xy term)
Bisectors of $ax^2+2hxy+by^2=0$: $\;\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
Perpendicular pair to given: replace $(a,h,b)\to(b,-h,a)$, giving $bx^2-2hxy+ay^2=0$
USE Bisectors of a pair are always perpendicular to each other (coeff of $x^2$ + coeff of $y^2=0$ in bisector eq).
Combined equation of lines $OP$, $OQ$ (chord intersections from origin):
$$ax^2+2hxy+by^2+2(gx+fy)\!\left(\frac{\ell x+my}{-n}\right)+c\!\left(\frac{\ell x+my}{-n}\right)^{\!2}=0$$
USE Find the pair of lines joining origin to intersection of curve and line — then use $a+b=0$ for perpendicularity condition (angle at origin = 90°).
🔄
Transformation of Axes
4 formulas▼
Shifting of Origin to $(\alpha,\beta)$
$$x = x'+\alpha,\quad y = y'+\beta \quad\Longleftrightarrow\quad x'=x-\alpha,\quad y'=y-\beta$$
Replace $x\to x+\alpha$ and $y\to y+\beta$ in the equation to get new equation.
USE Eliminate first-degree terms — shift origin to point of intersection / vertex of conic.
Rotation of Axes by Angle $\theta$ (anticlockwise)
$$x = x'\cos\theta - y'\sin\theta,\quad y = x'\sin\theta + y'\cos\theta$$
$$x' = x\cos\theta+y\sin\theta,\quad y' = -x\sin\theta+y\cos\theta$$
Table: New $x'$ uses $\cos\theta$ from old $x$, $\sin\theta$ from old $y$; New $y'$ uses $-\sin\theta$, $\cos\theta$.
USE Eliminate the xy term from a second-degree equation — choose θ so that coefficient of x'y' becomes zero (solve $\cot 2\theta=(a-b)/2h$).
▧
Linear Inequalities & Regions
3 formulas▼
Region Represented by $ax+by+c>0$
Test with a convenient point (e.g., origin):
If $a(0)+b(0)+c>0$ → origin is in the region $ax+by+c>0$.
The inequality represents the half-plane on one side of the line $ax+by+c=0$.
USE For feasibility region: shade intersection of all inequality half-planes — corner points give extrema of objective function.
Line Passing Inside a Triangle
Line $L=0$ passes inside $\triangle ABC$ if it separates at least one vertex from the other two.
i.e. $L(A)$ and $L(B)$ have opposite signs, or $L(A)$ and $L(C)$ have opposite signs.
USE Check signs of $L$ at each vertex — two on same side, one on other side → line passes through interior.
◉ Circle
Locus of equidistant points from centre · General form · Tangent · Chord · Power of a Point · Radical Axis
Core Formula
General / Definition
Theorem / Property
📐
Basic Circle Geometry — Key Theorems
9 theorems▼
Chord Perpendicularity
Line from centre bisects a chord $\;\Leftrightarrow\;$ it is perpendicular to the chord. (Two-way)
USE Perpendicular from centre to chord bisects it — use to find chord length given distance from centre.
Equal Chords & Distance
$AB = CD \;\Leftrightarrow\; OP = OQ$ (equal chords equidistant from centre; nearer chord is longer)
USE Greatest chord = diameter ($d=0$ from centre); shortest chord approaches diameter approaches the tangent.
USE Substitute $y=mx+c$ into circle, form quadratic in $x$, set discriminant $\ge 0$ for intersection.
Length of Chord
$$\ell = 2\sqrt{r^2-p^2} \quad \text{where }p=\text{perpendicular distance from centre to chord}$$
USE Use Pythagoras: half-chord $= \sqrt{r^2-p^2}$. Longest chord = diameter ($p=0$); approaches 0 as $p\to r$.
⟋
Tangent — All Forms
8 formulas▼
To $x^2+y^2=a^2$
Slope Form
$$y = mx \pm a\sqrt{1+m^2}$$
Point of contact: $\left(\frac{-a^2m}{c},\frac{a^2}{c}\right)$ where $c=\pm a\sqrt{1+m^2}$
USE Use when slope of tangent is known — gives two parallel tangents for each $m$.
Point Form
$$xx_1+yy_1 = a^2 \quad \text{at point }(x_1,y_1)\text{ on circle}$$
USE Replace $x^2\to xx_1$, $y^2\to yy_1$ in circle equation — the "T=0" substitution rule.
Parametric Form
$$x\cos\alpha + y\sin\alpha = a \quad \text{at }(a\cos\alpha, a\sin\alpha)$$
Intersection of tangents at $P(\alpha)$ and $Q(\beta)$:
$$\left(\frac{a\cos\!\tfrac{\alpha+\beta}{2}}{\cos\!\tfrac{\alpha-\beta}{2}},\;\frac{a\sin\!\tfrac{\alpha+\beta}{2}}{\cos\!\tfrac{\alpha-\beta}{2}}\right)$$
USE Parametric tangent simplest for angle-based problems — one clean formula instead of slope form.
Circumcircle of $\triangle PT_1T_2$: $(x-x_1)(x+g)+(y-y_1)(y+f)=0$
USE All four results derived from the geometry of $\triangle OPT$ (right angle at $T$) — $L=\sqrt{OP^2-R^2}$.
Chord with Given Midpoint ($T = S_1$)
Equation of Chord (Midpoint = $(x_1,y_1)$)
$$T = S_1$$
$$xx_1+yy_1+g(x+x_1)+f(y+y_1)+c = x_1^2+y_1^2+2gx_1+2fy_1+c$$
Slope of chord $= -\dfrac{x_1+g}{y_1+f}$ (perpendicular to line joining centre to midpoint)
USE Shortest chord through an interior point has its midpoint at that point — use $T=S_1$ for its equation.
↗↙
Pair of Tangents — $SS_1 = T^2$
3 formulas▼
Combined Equation of Pair of Tangents
$$SS_1 = T^2$$
where $S\equiv x^2+y^2+2gx+2fy+c$,\quad $S_1\equiv x_1^2+y_1^2+2gx_1+2fy_1+c$
$$T\equiv xx_1+yy_1+g(x+x_1)+f(y+y_1)+c$$
USE From external point $(x_1,y_1)$, $SS_1=T^2$ gives joint equation of both tangents as a pair of straight lines.
Chord of Contact from Pair of Tangents on $xy=c^2$
Tangent pair from origin to circle: $2x^2-3xy+y^2=0$ (example)
Angle $2\theta$ between tangents: use $\tan\theta = r/L$
Four co-normal points on $xy=c^2$: product of parameters $t_1t_2t_3t_4=-1$
USE $OA = r/\sin\theta$ where $\theta$ is half the angle between tangents — use angle formula to find distance $OA$.
🔵
Family of Circles
8 formulas▼
Through Intersection of Two Circles $S_1=0$, $S_2=0$
$$S_1 + \lambda S_2 = 0 \quad (\lambda\neq -1)$$
USE All circles through the two intersection points — find $\lambda$ from extra condition (passes through point, has given radius, etc.).
Through Intersection of Circle $S=0$ and Line $L=0$
$$S + \lambda L = 0$$
USE Smallest circle in this family: its centre lies on the line $L=0$ (set $-g/\lambda$ type centre on line and solve).
Through Two Given Points $(x_1,y_1)$ and $(x_2,y_2)$
USE Number of common tangents = 4, 3, 2, 1, or 0 based on relative position — memorise the boundary conditions.
External Centre of Similitude $P$
$$P \text{ divides }C_1C_2 \text{ externally in ratio }r_1:r_2$$
$$P = \left(\frac{r_1x_2-r_2x_1}{r_1-r_2},\;\frac{r_1y_2-r_2y_1}{r_1-r_2}\right)$$
(Direct common tangents pass through $P$)
USE Write equation of tangent from $P$ to either circle to find direct common tangents.
Internal Centre of Similitude $T$
$$T \text{ divides }C_1C_2 \text{ internally in ratio }r_1:r_2$$
$$T = \left(\frac{r_1x_2+r_2x_1}{r_1+r_2},\;\frac{r_1y_2+r_2y_1}{r_1+r_2}\right)$$
(Transverse common tangents pass through $T$)
USE Transverse tangents cross between the circles — internal division gives the crossing point.
USE Direct tangent uses $(r_1-r_2)^2$ (same direction radii); transverse uses $(r_1+r_2)^2$ (opposite radii).
Two Circles Touch Each Other — Condition
Externally: $d = r_1+r_2$ → point of contact divides $C_1C_2$ internally in $r_1:r_2$
Internally: $d = |r_1-r_2|$ → point of contact divides $C_1C_2$ externally in $r_1:r_2$
Circles $x^2+y^2+2ax+c^2=0$ and $x^2+y^2+2by+c^2=0$ touch each other if $\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{1}{c^2}$
USE For touching: distance between centres = sum (ext) or difference (int) of radii.
⊕
Radical Axis & Radical Centre
8 formulas▼
Radical Axis of $S_1=0$ and $S_2=0$
$$S_1 - S_2 = 0 \quad\Longrightarrow\quad 2(g_1-g_2)x+2(f_1-f_2)y+(c_1-c_2)=0$$
(Locus of equal power w.r.t. both circles)
USE First normalise both equations so coefficients of $x^2,y^2$ are each 1, then subtract directly.
Key Properties of Radical Axis
(a) Always $\perp$ to the line joining the centres
(b) If circles intersect → radical axis = common chord
(c) If circles touch → radical axis = common tangent at contact point
(d) Bisects each common external tangent
(e) Passes through midpoint of $C_1C_2$ only if $r_1=r_2$
(f) Concentric circles have no radical axis
USE Property (a): verify your radical axis is perpendicular to $C_1C_2$ as a quick sanity check.
Radical Centre
Common intersection of radical axes of three circles taken two at a time.
Equal tangent lengths from radical centre to all three circles.
USE Find radical axes of any two pairs → solve simultaneously → radical centre. Then $\sqrt{S_1}$ from this point = tangent to each circle.
⊗
Coaxial System & Orthogonal Circles
7 formulas▼
Coaxial System
Definition & Forms
Family with common radical axis = coaxial system.
Form 1 (radical axis $P=0$ and one circle $S=0$): $\quad S+\lambda P=0$
Form 2 (two circles $S_1=0$, $S_2=0$): $\quad S_1+\lambda(S_1-S_2)=0$
Simplest form: $\quad x^2+y^2+2gx+c=0$ ($g$ variable, $c$ constant)
USE Coaxial circles share a radical axis — all have centres on the same line perpendicular to the radical axis.
USE Tangents at intersection point of orthogonal circles pass through the other circle's centre.
Key Orthogonality Properties
(a) Centre of variable circle orthogonal to two fixed circles lies on their radical axis.
(b) Centre of circle orthogonal to three given circles = radical centre (if outside all three).
USE Use (a) to find locus of centre when orthogonality to two circles is the condition.
💡
Important Highlights & Quick Results
10 results▼
Circle on Focal Chord as Diameter
Circle drawn on any focal chord of a conic as diameter touches the corresponding directrix.
USE Shared property — parabola, ellipse, and hyperbola all satisfy this.
Chord of Contact from Directrix
Chord of contact from any point on the directrix passes through the corresponding focus.
USE Standard conic property — used in both circle and conic section proofs.
Angle in Semicircle
$\angle APB = 90°$ if $AB$ is diameter — use diameter form circle $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ directly.
USE Shortcut: if a triangle has sides such that $AB^2+BC^2=AC^2$, the circumcircle has $AC$ as diameter.
Circumcircle of Triangle $PT_1T_2$
$$(x-x_1)(x+g)+(y-y_1)(y+f)=0$$
(where $(x_1,y_1)$ is external point and $(-g,-f)$ is the centre)
USE Circumcircle of triangle formed by tangents and chord of contact — diameter is $OP$ (from external point to centre).
Two Circles: $x^2+y^2=ax$ and $x^2+y^2=c^2$ Touch iff $|a|=c$
Circle 1: centre $(a/2,0)$, radius $|a|/2$
Circle 2: centre $(0,0)$, radius $c$
Touch internally iff distance between centres = difference of radii: $|a|/2 = c - |a|/2 \Rightarrow |a|=c$
USE Common pattern — one circle passes through origin and touches the other.
Circle Through Origin: $x^2+y^2+\lambda y=0$ Touching y-axis at $(0,2)$
Family touching y-axis at $(0,k)$: $x^2+(y-k)^2+\lambda x=0$
Substitute other given point to find $\lambda$ → complete equation.
USE Quick family form — saves solving from general form. Don't forget to verify the given point satisfies the final equation.
Locus of Midpoint of Chord of Contact
Point $(x_1,y_1)$ on line $lx+my=1$ → chord of contact of $x^2+y^2=a^2$:
Midpoint $(h,k)$ satisfies: $h^2+k^2 = a^2(lh+mk)$
Locus: $x^2+y^2=a(lx+my)$ (a circle through origin)
USE Use $T=S_1$ for chord with midpoint $(h,k)$, then eliminate $(x_1,y_1)$ using the given line condition.
Common Chord of Two Circles
$$S_1-S_2=0 \quad \text{(radical axis = common chord when circles intersect)}$$
Length of common chord: use distance from either centre to the radical axis line, then $\ell = 2\sqrt{r^2-p^2}$.
USE Subtract the two circle equations (coefficients of $x^2,y^2$ must be 1 first) to get the common chord.