Circles

Sameer Chincholikar
#
Sameer Chincholikar
Director - Unacademy JEE/NEET
B.Tech, M.Tech - IIT-Roorkee
Taught 1 Million+ Students
10+ years Teaching experience
EdTech
Taught Patent Holder
1 Million+ Students
Certified by the IAPT
100+ Aspiring Teachers Mentored
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Circles
 Circle

It is the locus of a point which moves in a plane in such a way that

its distance from a fixed point is always constant.
Equation
of Circle
 Center Radius Form

1. x2 + y2 = a2,
where the centre is (0, 0) and the radius is a.

2. (x - x1)2 + (y - y1)2 = a2,

where the centre is (x1, y1) and the radius is a.
General Equation of a Circle
If the incentre of an equilateral triangle is (1, 1) and the equation of its
one side is 3x + 4y + 3 = 0, then the equation of the circumcircle of this
triangle is :

A. x2 + y2 - 2x - 2y - 14 = 0 B. x2 + y2 + 2x - 2y - 2 = 0

C. x + y + 2x - 2y + 2 = 0
2 2
D. x2 + y2 + 2x - 2y - 7 = 0

JEE Main 2015

A square is inscribed in the circle x2 + y2 - 6x + 8y - 103 = 0, with its
sides are parallel to the coordinates axes. Then the distance of the
vertex of this square which is nearest to the origin is:

√137 JEE Main 2019

A. 6 B.

C. √41 D. 13
A circle passes through the points (2, 3) and (4, 5). If its centre lies
on the line, y - 4x + 3 = 0, then its radius is equal to

A. √5 B. 1
JEE Main 2018
C. √2 D. 2
 Diameter form

The equation of a circle, when the end-points (x1, y1) and (x2, y2)
of a diameter are given is (x - x1) (x - x2) + (y - y1) (y - y2) = 0
Line 3x + 7y = 21 meets the axes at A and B. Find the equation of the
circle through O, A, B, where O is origin.
Parametric Equation of a Circle

A. If the equation of a circle be x2 + y2 = a2, its

parametric equations are
x = a cos θ, y = a sin θ.
where θ is a parameter.
Parametric Equation of a Circle
If the equation of a circle be (x - h)2 + (y - k)2 = a2,
B. its parametric equations are
x = h + a cosθ, y = k + a sinθ,
where θ is a parameter.
Find the parametric equations of circle x2 + y2 + 6x - 4y – 12 = 0
 Circle
[Under Special Conditions]
 Circles under special conditions

1. The equation of the circle touching x axis

(x - h)2 + (y - k)2 = k2

2. The equation of the circle touching y axis

(x - h)2 + (y - k)2 = h2

3. The equation of the circle touching both

axis (x ± r)2 + (y ± r)2 = r2

(r, r)
4. The equation of the circle passing
through origin x2 + y2 + 2gx + 2fy = 0.
The circle passing through (1, -2) and touching the axis of x at (3, 0) also
passes through the point

JEE Main 2013

The equation of the circle passing through (3, -6) and touching
both the axes is:
Length of
Intercept
Intercepts made on the axes by a circle

Let the circle be x2 + y2 + 2gx + 2fy + c = 0. Then

Circle(s) touching x-axis at a distance 3 from the origin
and having an intercept of length 2√7 on y-axis is (are)

JEE Adv. 2013

Position of Point
[w.r.t. a Circle]
Position of Point

Outside
Inside

On the circle
How many tangents can be drawn from the point (5/2, 1) to the
circumcircle of the triangle with vertices
(1, √3) (1, -√3), (3, -√3).
Position of Line
[w.r.t. a Circle]
 Position of Line

Non Intersecting

Tangent
Intersecting
NOTE:
Find the value of ‘k’ for which 3x - 4y + k = 0 will be tangent
to the circle x2 + y2 = 10x.
The equation of a circle which touches both axes and the lines
3x - 4y + 8 = 0 and whose centre lies in the third quadrant is

x2 + y2 + 2x + 4y - 2 = 0
A. B. x2 + y2 + 2x + 2y + 1 = 0

C. x2 + y2 + 4x + 4y + 4 = 0 D. x2 + y2 + 4x + 2y - 2 = 0
Equation of
Tangent
 Point form
The equation of the tangent to the circle
P(x1, y1)
x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1)

T
The tangent to the circle C1 : x2 + y2 – 2x – 1 = 0 at the point
(2, 1) cuts off a chord of length 4 from a circle C2 whose
centre is (3, –2). The radius of C2, is

A. √6 JEE Main 2018

B. 2
C. √2
D. 3
Parametric form
The equation of a tangent to the circle
x2 + y2 = r2 at (r cos 𝛉, r sin 𝛉).
The equations of the tangent to the circle x2 + y2 = a2 which
makes a triangle of area a2 with the coordinate axes, is
 Slope form

The equation of a tangent to the circle

x2 + y2 = r2 with slope m.
Important Result
➔ The coordinates of the point of contact are
Center not at origin?
Find the equation of tangents to the circle
x2 + y2 - 6x + 4y - 12 = 0, which are parallel to the line
4x - 3y + 10 = 0.
If a line, y = mx + c is a tangent to the circle
(x - 3)2 + y2 = 1 and it is perpendicular to line L1, where L1is the
tangent to the circle, x2 + y2 = 1 at the point (1/√2, 1/√2); then:

JEE Main 2020

Tangent from external point

P(x1, y1)
Find the equation of tangents to the circle
x2 + y2 - 6x + 8y = 0, from the point (0, 1).
Equation of
Normal
Normal
If a line is perpendicular to the tangent at the point of
contact then it is called a normal.

In case of circle we know that tangent is perpendicular to

radius at the point of contact.
Normal always passes through the centre of the circle

Y
P(x1, y1)

C(h, k)

X
O
The area of the triangle formed by the x-axis and the normal and the
tangent to the circle x2 + y2 = 4x at (1, √3) is
 Length of
Tangent
& Power of Point
Length of Tangent & Power of point

1. The length of the tangent from a point (x1, y1) to the circle
x2 + y2 = a2 is
Length of Tangent & Power of point

2. The length of the tangent from a point (x1, y1) to the circle
x2 + y2 + 2gx + 2fy + c = 0 is
 Power of a point with respect to a circle
➔ The power of a point P with respect to any circle is PA • PB.
➔ From the geometry, we can write PA • PB = PT2
➔ Thus, the power of a point is the square of the length of the
tangent to a circle from that point.
If the tangent at the point P on the circle x2 + y2 + 6x + 6y = 2 meets
a straight line 5x - 2y + 6 = 0 at a point Q on the y-axis, then the
length of PQ is
Chord of
Contact
Chord of Contact
➔ From any external point, two tangents can be drawn to a given
circle. The chord joining the points of contact of the two tangents
is called the chord of contact of tangents.

T=0
The chords of contact of the pair of tangents drawn from each
point on the line 2x + y = 4 to circle x2 + y2 = 1 always passes
through a fixed point (a , b). Then find the value of a/b.
The locus of the point of intersection of the tangents at the
extremities of a chord of the circle x2 + y2 = a2 which touches the
circle x2 + y2 = 2ax is
 Chord With
Given Midpoint
Chord with given Midpoint

T = S1
The locus of the midpoint of the chord of contact of tangents drawn from
points lying on the straight line 4x - 5y = 20 to the circle x2 + y2 = 9 is

JEE Adv. 2012

 Pair of
Tangents
Pair of tangents

S.S1 = T2
Find the equation of tangents to the circle x2 + y2 - 6x + 8y = 0, from the
point (0, 1).
Director
Circle
Director Circle
➔ It is the locus of the point of intersection of perpendicular
tangents.
Director Circle
➔ In case of circles, it is a concentric circle having radius √2 times
the radius of the original circle.

P(h, k)

B
The locus of a point of intersection of perpendicular tangents to the circle
x2 + y2 - 4x - 6y - 1 = 0
The tangents drawn from the origin to the circle
x2 + y2 + 2ax - 2by + b2 = 0 are perpendicular then a2 - b2 is
 Some
Important Results
Important Results
Let the tangents drawn from the origin to the circle,
x² + y² - 8x - 4y + 16 = 0 touch it at the points A and B. The (AB)² is equal to:

JEE Main 2020

Tangents drawn from the point P(1, 8) to the circle
x2 + y2 - 6x - 4y - 11 = 0 touch the circle at the points A and B.
The equation of the circumcircle of the triangle PAB is

JEE Adv. 2009

Relative Position
of two circles
Position Conditions Number of
Tangents

Non-Intersecting C1C2 > r1 + r2 4

Touching Externally C1C2 = r1 + r2 3

Intersecting |r1 - r2| < C1C2 < r1 + r2 2

Touching Internally C1C2 = |r1 - r2| 1

Circle inside Circle C1C2 < |r1 - r2| 0

The number of common tangents to the circles
x2 + y2 = 4 and x2 + y2 - 6x - 8y = 24 is
If the circles x² + y² - 16x - 20y + 164 = r² and
(x - 4)² + (y - 7)² = 36 intersect at two distinct points, then :

JEE Main 2019

Length and Equation
of
Common Tangents
Length of Tangents

1. Direct Common Tangent 2. Transverse Common

Tangent
Find the equation of common tangents to the circles
x2 + y2 - 12x - 8y + 36 = 0 and x2 + y2 - 4x - 2y + 4 = 0
touching the circles in the distinct points.
Orthogonality
of two circles
 Orthogonal Circles
If the angle between the two circles is 900, then the circles are
said to be Orthogonal Circles.
 Condition for Orthogonality
Consider

and
These circles will be orthogonal if
A circle S passes through the point (0, 1) and is orthogonal to the circles
(x - 1)2 + y2 = 16 and x2 + y2 = 1. Then

JEE Adv. 2014

If the circle passes through the point (a, b) and cuts the circle
x2 + y2 = k2 orthogonally, then the equation of the locus of its centre is
Two circles with equal radii are intersecting at the points (0, 1) and (0, -1).
The tangent at the point (0, 1) to one of the circles passes through the
centre of the other circle. Then the distance between the centres of these
circles is :

JEE Main 2019

Radical
Axis
Radical axis

The radical axis of two circles is the locus of a point which moves in a
plane in such a way that the lengths of the tangents drawn from it to
the two circles is equal.
Radical axis

S - S’ = 0
Properties of the Radical Axis

1. For intersecting circles, the common chord and the

radical axis are identical.
Properties of the Radical Axis

2. If the two circles touch each other externally or

internally, the common tangents and the radical axis
are identical
Properties of the Radical Axis
3. The radical axis is perpendicular to the straight lines which
joins the centres of the circles.The radical axis is perpendicular
to the straight lines which joins the centres of the circles.
Properties of the Radical Axis
4 The radical axis bisects common tangents of two circles.
.
Properties of the Radical Axis
5. Only concurrent circles do not have a radical axis.
Properties of the Radical Axis
6. If two circles cut a third circle orthogonally, the radical axis
of the two circles will pass through the centre of the third
circle.
Let the circles S ≡ x2 + y2 - 12 = 0 and S’ ≡ x2 + y2 - 5x + 3y -2 = 0
intersect at points P and Q. Tangents are drawn to the circle S at
points P and Q. Then point of intersection of tangents is:
The length of common chord of the circles x2 + y2 = 12 and
x2 + y2 - 4x + 3y - 2 = 0, is
The common tangent to the circle x2 + y2 = 4 and
x2 + y2 + 6x + 8y - 24 = 0 also passes through the point:

JEE Main 2019

Radical
Center
Radical Center
The radical axis of three circles taken in pairs are concurrent at a
point called radical center.
NOTE: Circle orthogonal to three circles
The radical centre of the three given circles will be the centre of a fourth
circle, which cuts all the three circles orthogonally and the radius of the
fourth circle is the length of the tangent drawn from the radical centre of
the three given circles to any of these circles.
Radical Center

The radical axis of three circles taken in pairs are concurrent at a

point called radical center.
NOTE: Circle orthogonal to three circles
The radical centre of the three given circles will be the centre of a fourth
circle, which cuts all the three circles orthogonally and the radius of the
fourth circle is the length of the tangent drawn from the radical centre of
the three given circles to any of these circles.
Family of
Circles
Family of circles
1. The equation of the family of circles passing through the
points of intersection of two circles
S = 0 & S’ = 0 is: S + λ S’ = 0
(λ ≠ -1 provided the coefficient of x2 & y2 in S & S’ are same).
Family of circles
2. The equation of the family of circles passing through the
point of intersection of a circle S = 0 & a line L = 0 is given
by S + λL = 0.
Family of circles
3. The equation of a family of circles passing through two
given points (x1, y1) & (x2, y2) can be written in the form:
Family of circles
4. The equation of a family of circles touching a fixed line
y - y1 = m(x - x1) at the fixed point (x1, y1) is
(x - x1)2 + (y - y1)2 + λ [(y - y1)- m(x - x1)] = 0,
The circle passing through the intersection of the circles x2 + y2 - 6x = 0
and x2 + y2 - 4y = 0, having its centre on the line, 2x - 3y + 12 = 0, also
passes through the point:

A. (-1, 3) B. (-3, 6) C. (-3, 1) D. (1, -3)

JEE Main 2020
If y + 3x = 0 is the equation of a chord of the circle, x2 + y2 - 30x = 0, then
the equation of the circle with this chord as diameter is:

A. x2 + y2 + 3x + 9y = 0 B. x2 + y2 + 3x - 9y = 0
JEE Main 2015
C. x2 + y2 - 3x - 9y = 0 D. x2 + y2 - 3x + 9y = 0
The circle passing through the point (-1, 0) and touching the
y-axis at (0, 2) also passes through the point.

JEE Adv. 2011

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