Involving Two or More Circles

Q: How do you find whether two circles intersect, touch or don't meet?

Compute the distance d between the centres and compare with r_1 + r_2 and |r_1 - r_2|. If d \gt r_1 + r_2 they are separated, d = r_1 + r_2 externally touch, |r_1 - r_2| \lt d \lt r_1 + r_2 they intersect at two points, d = |r_1 - r_2| internally touch, and d \lt |r_1 - r_2| one lies inside the other with no contact.

Q: What is the equation of the common chord of two circles?

If the two circles are S_1 = 0 and S_2 = 0 (both with the coefficient of x^2 and y^2 equal to 1), the common chord is S_1 - S_2 = 0. This is a straight line - it is also the radical axis of the two circles.

Q: How do you compute the length of the direct and transverse common tangents?

With d = distance between centres and r_1, r_2 = radii, L_{\text{ext}} = \sqrt{d^2 - (r_1 - r_2)^2} for direct tangents and L_{\text{int}} = \sqrt{d^2 - (r_1 + r_2)^2} for transverse tangents. Transverse only exists when d \ge r_1 + r_2.

Q: What is the condition for two circles to intersect orthogonally?

Two circles x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0 and x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0 are orthogonal if and only if 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2. Geometrically, (C_1 C_2)^2 = r_1^2 + r_2^2.

Q: What is the radical axis of two circles?

The radical axis is the locus of points whose tangent lengths to the two circles are equal. Its equation is S_1 - S_2 = 0 and it is always perpendicular to the line joining the centres. If the circles intersect, the radical axis is their common chord.

Q: What is the radical centre of three circles?

Taking the three circles pairwise gives three radical axes, and they always meet at a single point called the radical centre. Tangent lengths from the radical centre to all three circles are equal. It is useful for setting up equal-tangent problems.

Q: What is a coaxial system of circles?

A family of circles all of which share the same radical axis is called a coaxial system. In canonical form, every circle in the system can be written as x^2 + y^2 + 2gx + c = 0 with g varying and c fixed. This is a JEE Advanced concept and often appears alongside limiting points.

Q: What are the limiting points of a coaxial system?

The limiting points are the point circles in a coaxial system - obtained by setting radius = 0, which gives g^2 = c, so g = \pm\sqrt{c} (when c \gt 0). Every circle through the two limiting points is orthogonal to every circle of the coaxial system.

Q: How do you find the equation of a circle passing through the intersection of two given circles?

Use the family S_1 + \lambda S_2 = 0, where \lambda is a real parameter with \lambda \ne -1. Then apply the extra condition (radius, a third point, etc.) to solve for \lambda.

Q: How do you find the circle circumscribing a triangle whose sides are given?

Use L_1 L_2 + \lambda L_2 L_3 + \mu L_1 L_3 = 0, then impose two conditions: coefficient of x^2 equals coefficient of y^2, and coefficient of xy = 0. This gives two linear equations in \lambda, \mu that fix the circle uniquely.

MathsCirclesFor JEE aspirants

Two circles can be positioned five ways relative to each other, and the count of common tangents tells you which: separated (4), externally touching (3), intersecting (2), internally touching (1), one inside the other (0). Two circles $S_{1} = 0$ and $S_{2} = 0$ intersect orthogonally when $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ . This concept covers common tangents, common chord, angle of intersection, orthogonality, radical axis and radical centre, family of circles, and the JEE Advanced topic of coaxial systems and limiting points.

Key Formulas - Quick Reference

Common chord of two intersecting circles: $S_{1} - S_{2} = 0$
Length of common chord of $S_{1} = 0$ and $S_{2} = 0$ : $2 r_{1}^{2} - p_{1}^{2}$ where $p_{1}$ = distance from centre of $S_{1}$ to the line $S_{1} - S_{2} = 0$
Length of direct (external) common tangent: $L_{ext} = d^{2} - (r_{1} - r_{2})^{2}$
Length of transverse (internal) common tangent: $L_{int} = d^{2} - (r_{1} + r_{2})^{2}$
Direct common tangents meet on the line joining the centres, dividing it externally in ratio $r_{1} : r_{2}$
Transverse common tangents meet on the line joining the centres, dividing it internally in ratio $r_{1} : r_{2}$
Orthogonality condition for two circles: $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
Radical axis of two circles: $S_{1} - S_{2} = 0$ (always perpendicular to the line joining the centres)
Family of circles through intersection of $S_{1} = 0$ and $S_{2} = 0$ : $S_{1} + λ S_{2} = 0, λ \neq = - 1$
Family of circles through intersection of circle $S = 0$ and line $L = 0$ : $S + λ L = 0$
Family touching a fixed line $y - y_{1} = m (x - x_{1})$ at fixed point $(x_{1}, y_{1})$ : $(x - x_{1})^{2} + (y - y_{1})^{2} + λ [y - y_{1} - m (x - x_{1})] = 0$

1. Relative Position of Two Circles

Given two circles with centres $C_{1}$ , $C_{2}$ and radii $r_{1}$ , $r_{2}$ , let $d = C_{1} C_{2}$ be the distance between the centres. There are exactly five cases, distinguished by comparing $d$ with $r_{1} + r_{2}$ and $∣ r_{1} - r_{2} ∣$ . Every question about two circles starts here, so it is worth being able to draw all five from memory.

Figure 1: The five relative positions of two circles, with every common tangent drawn. Which case you are in is decided by comparing

d

with

r_{1} + r_{2}

and

∣ r_{1} - r_{2} ∣

Case	Condition	Common tangents
Circles are separated (one outside the other)	$d > r_{1} + r_{2}$	4 (2 direct + 2 transverse)
Externally touching	$d = r_{1} + r_{2}$	3 (2 direct + 1 transverse)
Intersecting at two points	$∣ r_{1} - r_{2} ∣ < d < r_{1} + r_{2}$	2 (both direct)
Internally touching	$d = ∣ r_{1} - r_{2} ∣$	1 (direct)
One circle inside the other (no touching)	$d < ∣ r_{1} - r_{2} ∣$	0

Reading the table backwards. Problems often give you the number of common tangents and ask for a condition on a parameter. Convert the tangent count straight into the corresponding line of the table, then solve the resulting equation or inequality in

d

Solved Example 1

Examine if the circles

x^{2} + y^{2} - 2 x - 4 y = 0

and

x^{2} + y^{2} - 8 y - 4 = 0

touch each other. If so, is the contact external or internal?

Solution:

First circle: centre $A = (1, 2)$ , radius $r_{1} = 1 + 4 = 5$ .

Second circle: centre $B = (0, 4)$ , radius $r_{2} = 0 + 16 + 4 = 20 = 25$ .

$A B = 1 + 4 = 5, r_{1} + r_{2} = 35, ∣ r_{1} - r_{2} ∣ = 5 .$

Since $A B = ∣ r_{1} - r_{2} ∣$ , the circles touch each other internally.

2. Common Chord of Two Intersecting Circles

If two circles $S_{1} = 0$ and $S_{2} = 0$ intersect at two points $A$ and $B$ , then $A$ and $B$ satisfy both $S_{1} = 0$ and $S_{2} = 0$ , and hence also satisfy $S_{1} - S_{2} = 0$ , which is a straight line. So the equation of the common chord is

$S_{1} - S_{2} = 0.$

Length of the common chord. Let $p_{1}$ be the perpendicular distance from the centre of $S_{1} = 0$ to the line $S_{1} - S_{2} = 0$ . Then

$length of common chord = 2 r_{1}^{2} - p_{1}^{2} .$

(You could equally use $p_{2}$ and $r_{2}$ from the other circle - the answer is the same.)

Both circles must be normalised first.

S_{1} - S_{2} = 0

is a straight line only because the

x^{2}

and

y^{2}

terms cancel, and they cancel only when the coefficient of

x^{2}

(and of

y^{2}

) is

1

in both equations. Divide through before subtracting.

3. Common Tangents to Two Circles

A common tangent touches both circles. It is called a direct (or external) common tangent if both centres lie on the same side of it, and a transverse (or internal) common tangent if the centres lie on opposite sides, so that the tangent crosses the segment joining them.

Figure 2: Two separated circles have four common tangents. The two direct tangents do not cross the line of centres between the circles and meet at

P_{ext}

, which divides

C_{1} C_{2}

externally in the ratio

r_{1} : r_{2}

. The two transverse tangents cross between the circles and meet at

P_{int}

, which divides

C_{1} C_{2}

internally in the same ratio.

Where the common tangents meet the line of centres.

The two direct common tangents meet at a point $P_{ext}$ that divides $C_{1} C_{2}$ externally in ratio $r_{1} : r_{2}$ .
The two transverse common tangents meet at a point $P_{int}$ that divides $C_{1} C_{2}$ internally in ratio $r_{1} : r_{2}$ .

Finding the equations of the common tangents

Write down the centres and radii of both circles, normalising each equation so that the coefficient of $x^{2}$ and $y^{2}$ is $1$ .
Locate $P_{ext}$ (for the direct pair) or $P_{int}$ (for the transverse pair) using the section formula on $C_{1} C_{2}$ in the ratio $r_{1} : r_{2}$ .
Write the pair of tangents from that point to either circle using $S S_{1} = T^{2}$ .
Factorise the resulting second degree equation into two linear factors; these are the two required tangents.

Lengths of common tangents

If $d = C_{1} C_{2}$ is the distance between centres and $r_{1}$ , $r_{2}$ are the radii, then

$L_{ext} = d^{2} - (r_{1} - r_{2})^{2}, L_{int} = d^{2} - (r_{1} + r_{2})^{2} .$

Here $L_{ext}$ and $L_{int}$ mean the distance between the two points of contact, not the length of the whole line. Since $(r_{1} + r_{2})^{2} \geq (r_{1} - r_{2})^{2}$ , the transverse common tangent is always shorter than the direct one.

4. Angle of Intersection and Orthogonality of Two Circles

If two circles intersect, the angle between them at a point of intersection is defined as the angle between their tangents at that point, which is the same as the angle between their radii drawn to that point.

Figure 3: Two circles cut orthogonally when the tangents at a point of intersection are perpendicular, that is when

C_{1} P ⊥ C_{2} P

. Each radius produced is the tangent to the other circle, so the tangent to one circle at

P

passes through the centre of the other. This gives

(C_{1} C_{2})^{2} = r_{1}^{2} + r_{2}^{2}

, and hence

2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}

Orthogonal intersection. Two circles are orthogonal if they intersect at

9 0^{\circ}

. This happens when the tangent to one at the point of intersection passes through the centre of the other, i.e. when

C_{1} P ⊥ C_{2} P

, which gives

(C_{1} C_{2})^{2} = r_{1}^{2} + r_{2}^{2}

Orthogonality condition

For circles $S_{1} \equiv x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $S_{2} \equiv x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ , orthogonality means

$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2} .$

Derivation. $(C_{1} C_{2})^{2} = (g_{1} - g_{2})^{2} + (f_{1} - f_{2})^{2}$ and $r_{1}^{2} + r_{2}^{2} = (g_{1}^{2} + f_{1}^{2} - c_{1}) + (g_{2}^{2} + f_{2}^{2} - c_{2})$ . Setting them equal and cancelling like terms gives the condition.

General angle of intersection

If the angle between the two circles is $θ$ , applying the cosine rule to triangle $C_{1} P C_{2}$ with $C_{1} P = r_{1}$ , $C_{2} P = r_{2}$ and $C_{1} C_{2} = d$ gives

$cos θ = \frac{r _{1}^{2} + r _{2}^{2} - d ^{2}}{2 r _{1} r _{2}} .$

Putting $θ = 9 0^{\circ}$ recovers $d^{2} = r_{1}^{2} + r_{2}^{2}$ , which is the orthogonality condition above.

Solved Example 2

Obtain the equation of the circle orthogonal to both

x^{2} + y^{2} + 3 x - 5 y + 6 = 0

and

4 x^{2} + 4 y^{2} - 28 x + 29 = 0

, and whose centre lies on the line

3 x + 4 y + 1 = 0

Solution:

Normalise the second circle: $x^{2} + y^{2} - 7 x + \frac{29}{4} = 0.$

Let the required circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Orthogonal to first circle: $2 g \cdot \frac{3}{2} + 2 f \cdot (- \frac{5}{2}) = c + 6 \Rightarrow 3 g - 5 f = c + 6$ ...(i)

Orthogonal to second circle: $2 g \cdot (- \frac{7}{2}) + 0 = c + \frac{29}{4} \Rightarrow - 7 g = c + \frac{29}{4}$ ...(ii)

(i) $-$ (ii): $10 g - 5 f = - \frac{5}{4} \Rightarrow 40 g - 20 f = - 5$ ...(iii)

Centre $(- g, - f)$ lies on $3 x + 4 y + 1 = 0$ : $- 3 g - 4 f = - 1$ ...(iv)

Solving (iii), (iv): $g = 0, f = \frac{1}{4}$ . Then $c = - \frac{29}{4} .$

Required circle: $4 (x^{2} + y^{2}) + 2 y - 29 = 0.$

5. Radical Axis and Radical Centre

Recall that the power of a point $P$ with respect to a circle equals $S_{1}$ (the value of the circle expression at $P$ ), and that for a point outside the circle the length of the tangent from it is $S_{1}$ . The radical axis of two circles is the locus of points whose powers with respect to the two circles are equal, equivalently the locus of points from which the two tangent lengths are equal.

Setting the two powers equal gives $S_{1} - S_{2} = 0$ , i.e.

$2 (g_{1} - g_{2}) x + 2 (f_{1} - f_{2}) y + (c_{1} - c_{2}) = 0.$

This is a straight line, and its normal vector $(2 (g_{1} - g_{2}), 2 (f_{1} - f_{2}))$ is parallel to $C_{1} C_{2}$ , which is why the radical axis is always perpendicular to the line of centres.

Figure 4: The radical axis is the locus of points from which the two tangent lengths are equal, so

P T_{1} = P T_{2} = L

for every point

P

on it. Its equation is

S_{1} - S_{2} = 0

, and it is always perpendicular to the line joining the centres.

Properties of the radical axis

If the two circles intersect, the radical axis is the common chord.
If they touch each other, the radical axis is the common tangent at the point of contact.
The radical axis is always perpendicular to the line joining the centres of the two circles.
It passes through the midpoint of the line joining the centres only if the two circles have equal radii.
It bisects any common tangent between the two circles.
A system of circles, every two of which have the same radical axis, is called a coaxial system.
Pairs of concentric circles do not have a radical axis (their radical "axis" is the whole plane if they are equal, or empty if unequal - the formula breaks down).

Radical centre of three circles

Given three circles, take them two at a time to get three radical axes. These three lines are concurrent at a single point, called the radical centre. The tangent lengths from the radical centre to all three circles are equal.

Why they are concurrent. Let the radical axes of the pairs

(S_{1}, S_{2})

and

(S_{2}, S_{3})

meet at

R

. Then the power of

R

is the same for

S_{1}

and

S_{2}

, and the same for

S_{2}

and

S_{3}

; hence it is the same for

S_{1}

and

S_{3}

, so

R

also lies on the third radical axis.

Solved Example 3

Find the point from which the tangents to

x^{2} + y^{2} - 4 x + 7 = 0

2 x^{2} + 2 y^{2} - 3 x + 5 y + 9 = 0

and

x^{2} + y^{2} + y = 0

are equal in length, and find that length.

Solution:

Normalise the middle circle: $x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2} = 0.$ Call the three circles $S_{1}, S_{2}, S_{3}$ .

Radical axis of $S_{1}, S_{2}$ : $S_{1} - S_{2} = - 4 x + 7 - (- \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2}) = 0 \Rightarrow - \frac{5}{2} x - \frac{5}{2} y + \frac{5}{2} = 0 \Rightarrow x + y = 1.$

Radical axis of $S_{2}, S_{3}$ : $S_{2} - S_{3} = - \frac{3}{2} x + \frac{5}{2} y + \frac{9}{2} - y = 0 \Rightarrow - \frac{3}{2} x + \frac{3}{2} y + \frac{9}{2} = 0 \Rightarrow x - y = 3.$

Solving $x + y = 1, x - y = 3$ : $x = 2, y = - 1$ .

Length of tangent from $(2, - 1)$ to $S_{1} = 0$ : $4 + 1 - 8 + 7 = 4 = 2.$

Radical centre $(2, - 1)$ ; equal tangent length $= 2$ .

JEE Advanced

Coaxial System and Limiting Points

A family of circles that share a common radical axis is called a coaxial system. If we take the radical axis as the y-axis and the line of centres as the x-axis, every circle in the system has the form

$x^{2} + y^{2} + 2 g x + c = 0,$

where $g$ varies and $c$ is a fixed constant (the same for all circles in the system).

The limiting points of the system are the point circles that belong to it, obtained by setting the radius $= 0$ : $g^{2} = c$ , so $g = \pm c$ (assuming $c > 0$ ). The two limiting points are $(\pm c, 0)$ ; if $c < 0$ , the system has no real limiting points.

Every circle passing through the two limiting points is orthogonal to every circle of the coaxial system. Limiting points and coaxial systems appear regularly in JEE Advanced problems on families of circles.

6. Family of Circles

Below are the standard families of circles - each is a one-parameter (or two-parameter) family satisfying a given geometric constraint. In each, $λ$ (and $μ$ ) is a real parameter, fixed at the end by whatever extra condition the problem supplies.

(a) Family through intersection of two circles

The family of circles through the intersection points of $S_{1} = 0$ and $S_{2} = 0$ is

$S_{1} + λ S_{2} = 0, λ \neq = - 1$

(the $λ = - 1$ value gives the radical axis, not a circle).

(b) Family through intersection of a circle and a line

The family of circles through the intersection of $S = 0$ and line $L = 0$ is

$S + λ L = 0.$

(c) Family passing through two given points

The family of circles through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) + λ x y 1 x_{1} y_{1} 1 x_{2} y_{2} 1 = 0.$

The first two terms give the "diameter form" circle (smallest through the two points); the determinant is the line through them.

(d) Family touching a fixed line at a fixed point

The family of circles touching the line $y - y_{1} = m (x - x_{1})$ at the point $(x_{1}, y_{1})$ is

$(x - x_{1})^{2} + (y - y_{1})^{2} + λ [y - y_{1} - m (x - x_{1})] = 0.$

(e) Family circumscribing a triangle

Let the sides of a triangle be $L_{1} = 0$ , $L_{2} = 0$ , $L_{3} = 0$ . The family of circles circumscribing this triangle is

$L_{1} L_{2} + λ L_{2} L_{3} + μ L_{3} L_{1} = 0,$

with $λ$ , $μ$ chosen so that (i) coefficient of $x^{2}$ equals coefficient of $y^{2}$ , and (ii) coefficient of $x y$ is $0$ . These two conditions fix $λ$ and $μ$ , giving a unique circle.

(f) Family circumscribing a quadrilateral

If the sides of a quadrilateral (in order) are $L_{1} = 0$ , $L_{2} = 0$ , $L_{3} = 0$ , $L_{4} = 0$ , then the circle circumscribing the quadrilateral (when it exists - i.e. the quadrilateral is cyclic) has the form

$u L_{1} L_{3} + λ L_{2} L_{4} = 0,$

with $u$ and $λ$ chosen so that coefficient of $x^{2}$ $=$ coefficient of $y^{2}$ and coefficient of $x y = 0$ .

Solved Example 4

Find the equations of circles passing through the intersection points of

x^{2} + y^{2} - 2 x - 4 y - 4 = 0

and

x^{2} + y^{2} - 10 x - 12 y + 40 = 0

, and whose radius is

4

Solution:

Any circle through the intersection: $S_{1} + λ S_{2} = 0$ :

$(1 + λ) (x^{2} + y^{2}) - 2 (1 + 5 λ) x - 2 (2 + 6 λ) y + (- 4 + 40 λ) = 0.$

Divide by $(1 + λ)$ : $x^{2} + y^{2} - \frac{2 ( 1 + 5 λ )}{1 + λ} x - \frac{2 ( 2 + 6 λ )}{1 + λ} y + \frac{40 λ - 4}{1 + λ} = 0.$

Centre: $(\frac{1 + 5 λ}{1 + λ}, \frac{2 + 6 λ}{1 + λ})$ ; radius $^{2} = g^{2} + f^{2} - c$ .

Setting radius $= 4$ and simplifying yields $5 λ^{2} - 34 λ - 7 = 0 \Rightarrow λ = 7$ or $λ = - \frac{1}{5} .$

$λ = 7$ : $2 x^{2} + 2 y^{2} - 18 x - 22 y + 69 = 0.$

$λ = - \frac{1}{5}$ : $x^{2} + y^{2} - 2 y - 15 = 0.$

Solved Example 5

Find the equation of the circle passing through the point

(2, 1)

and touching the line

x + 2 y - 1 = 0

at the point

(3, - 1)

Solution:

Use family (d) with fixed point $(3, - 1)$ and line $x + 2 y - 1 = 0$ :

$(x - 3)^{2} + (y + 1)^{2} + λ (x + 2 y - 1) = 0.$

It passes through $(2, 1)$ : $1 + 4 + λ (2 + 2 - 1) = 0 \Rightarrow λ = - \frac{5}{3} .$

Required circle: $(x - 3)^{2} + (y + 1)^{2} - \frac{5}{3} (x + 2 y - 1) = 0$ , i.e. $3 x^{2} + 3 y^{2} - 23 x - 4 y + 35 = 0.$

Solved Example 6

Find the equation of the circle circumscribing the triangle whose sides are

3 x - y - 9 = 0

5 x - 3 y - 23 = 0

and

x + y - 3 = 0

Solution:

Family: $L_{1} L_{2} + λ L_{2} L_{3} + μ L_{1} L_{3} = 0$ with $L_{1} = 3 x - y - 9, L_{2} = 5 x - 3 y - 23, L_{3} = x + y - 3.$

After expanding, imposing (i) coefficient of $x^{2} =$ coefficient of $y^{2}$ and (ii) coefficient of $x y = 0$ gives

$2 λ + μ + 3 = 0$ ...(i), $λ + μ - 7 = 0$ ...(ii).

Solving: $λ = - 10, μ = 17.$

Substituting back and simplifying yields $2 x^{2} + 2 y^{2} - 5 x + 11 y - 3 = 0.$

Common Mistakes to Avoid

Watch out

Not normalising circles before applying orthogonality. The condition $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ requires the coefficient of $x^{2}$ (and $y^{2}$ ) to be $1$ in both circles. Divide first if it isn't.
Same mistake for the radical axis / common chord. $S_{1} - S_{2} = 0$ works only when both circles have $x^{2} + y^{2}$ (coefficient $1$ ) at the front.
Wrong side of the section formula. Direct common tangents meet externally in ratio $r_{1} : r_{2}$ ; transverse meet internally. Mixing them up flips the geometry.
Forgetting $λ \neq = - 1$ in family (a). $S_{1} + λ S_{2}$ at $λ = - 1$ is the radical axis (a line), not a circle. Similarly, if you're told the required circle has some property and $λ = - 1$ pops out, re-examine.
Applying the length formula for common tangents when circles overlap. $L_{int}^{2} = d^{2} - (r_{1} + r_{2})^{2}$ is meaningful only when $d \geq r_{1} + r_{2}$ (i.e. the transverse tangent exists). Otherwise the formula gives a negative number under the square root.
Assuming the radical axis always exists as a line. Two concentric circles have no radical axis (the algebra gives $c_{1} = c_{2}$ or a contradiction).
Losing track of $λ$ and $μ$ in the triangle-circumscribing family. There are two conditions to impose (both come from "this must be a circle"), giving two parameters uniquely.

Frequently Asked Questions

How do you find whether two circles intersect, touch or don't meet?

Compute the distance $d$ between the centres and compare with $r_{1} + r_{2}$ and $∣ r_{1} - r_{2} ∣$ . If $d > r_{1} + r_{2}$ they are separated, $d = r_{1} + r_{2}$ externally touch, $∣ r_{1} - r_{2} ∣ < d < r_{1} + r_{2}$ they intersect at two points, $d = ∣ r_{1} - r_{2} ∣$ internally touch, and $d < ∣ r_{1} - r_{2} ∣$ one lies inside the other with no contact.

What is the equation of the common chord of two circles?

If the two circles are $S_{1} = 0$ and $S_{2} = 0$ (both with the coefficient of $x^{2}$ and $y^{2}$ equal to $1$ ), the common chord is $S_{1} - S_{2} = 0$ . This is a straight line - it is also the radical axis of the two circles.

How do you compute the length of the direct and transverse common tangents?

With $d$ = distance between centres and $r_{1}$ , $r_{2}$ = radii, $L_{ext} = d^{2} - (r_{1} - r_{2})^{2}$ for direct tangents and $L_{int} = d^{2} - (r_{1} + r_{2})^{2}$ for transverse tangents. Transverse only exists when $d \geq r_{1} + r_{2}$ .

What is the condition for two circles to intersect orthogonally?

Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ are orthogonal if and only if $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ . Geometrically, $(C_{1} C_{2})^{2} = r_{1}^{2} + r_{2}^{2}$ .

What is the radical axis of two circles?

The radical axis is the locus of points whose tangent lengths to the two circles are equal. Its equation is $S_{1} - S_{2} = 0$ and it is always perpendicular to the line joining the centres. If the circles intersect, the radical axis is their common chord.

What is the radical centre of three circles?

Taking the three circles pairwise gives three radical axes, and they always meet at a single point called the radical centre. Tangent lengths from the radical centre to all three circles are equal. It is useful for setting up equal-tangent problems.

What is a coaxial system of circles?

A family of circles all of which share the same radical axis is called a coaxial system. In canonical form, every circle in the system can be written as $x^{2} + y^{2} + 2 g x + c = 0$ with $g$ varying and $c$ fixed. This is a JEE Advanced concept and often appears alongside limiting points.

What are the limiting points of a coaxial system?

The limiting points are the point circles in a coaxial system - obtained by setting radius $= 0$ , which gives $g^{2} = c$ , so $g = \pm c$ (when $c > 0$ ). Every circle through the two limiting points is orthogonal to every circle of the coaxial system.

How do you find the equation of a circle passing through the intersection of two given circles?

Use the family $S_{1} + λ S_{2} = 0$ , where $λ$ is a real parameter with $λ \neq = - 1$ . Then apply the extra condition (radius, a third point, etc.) to solve for $λ$ .

How do you find the circle circumscribing a triangle whose sides are given?

Use $L_{1} L_{2} + λ L_{2} L_{3} + μ L_{1} L_{3} = 0$ , then impose two conditions: coefficient of $x^{2}$ equals coefficient of $y^{2}$ , and coefficient of $x y = 0$ . This gives two linear equations in $λ, μ$ that fix the circle uniquely.

Previous year questions on Involving Two or More Circles

20 questions from past papers, each with a step-by-step solution.

Show all 20 questions

Ready to master Circles?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test