Circles.pdf

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11th / 9, 10
Plus ICONIC
11th / 9, 10
SAMEERLIVE
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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.
Circle

1. x2 + y2 = a2,
where the centre is (0, 0) and the radius is a.
Center Radius Form
2. (x - x1)2 + (y - y1)2 = a2,
where the centre is (x1, y1) and the radius is a.

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. B.
C. D. x2 + y2 + 2x - 2y - 7 = 0
x2 + y2 + 2x - 2y + 2 = 0
x2 + y2 + 2x - 2y - 2 = 0
x2 + y2 - 2x - 2y - 14 = 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:
A. B.
C. D. 13
√41
√137
6
JEE Main 2019

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. B.
C. D. 2
√2
1
√5
JEE Main 2018

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
Diameter form

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.

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

Parametric Equation of a Circle
B.
If the equation of a circle be (x - h)2 + (y - k)2 = a2,
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]

1. The equation of the circle touching x axis
(x - h)2 + (y - k)2 = k2
Circles under special conditions
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
4. The equation of the circle passing
through origin x2 + y2 + 2gx + 2fy = 0.
(r, r)

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:

Let the circle be x2 + y2 + 2gx + 2fy + c = 0. Then
Intercepts made on the axes by a circle

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]

Inside
Outside
On the circle
Position of Point

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]

Intersecting
Position of Line
Non Intersecting
Tangent

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

T
P(x1, y1)
The equation of the tangent to the circle
x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1)
Point form

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
B. 2
C. √2
D. 3
JEE Main 2018

The equation of a tangent to the circle
x2 + y2 = r2 at (r cos 𝛉, r sin 𝛉).
Parametric form

The equations of the tangent to the circle x2 + y2 = a2 which
makes a triangle of area a2 with the coordinate axes, is

The equation of a tangent to the circle
x2 + y2 = r2 with slope m.
Slope form

➔ The coordinates of the point of contact are
Important Result

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

T
P(x1, y1)
Tangent from external point

Find the equation of tangents to the circle
x2 + y2 - 6x + 8y = 0, from the point (0, 1).

In case of circle we know that tangent is perpendicular to
radius at the point of contact.
If a line is perpendicular to the tangent at the point of
contact then it is called a normal.
Normal

P(x1, y1)
C(h, k)
O
X
Y
Normal always passes through the centre of the circle

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

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
Length of Tangent & Power of point

➔ 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.
Power of a point with respect to a circle

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

➔ 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
Chord of Contact

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

T = S1
Chord with given Midpoint

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

Find the equation of tangents to the circle x2 + y2 - 6x + 8y = 0, from the
point (0, 1).

➔ 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
A
Director Circle

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

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.

If the angle between the two circles is 900, then the circles are
said to be Orthogonal Circles.
Orthogonal Circles

Consider
These circles will be orthogonal if
and
Condition for Orthogonality

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

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

For intersecting circles, the common chord and the
radical axis are identical.
1.
Properties of the Radical Axis

If the two circles touch each other externally or
internally, the common tangents and the radical axis
are identical
2.

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.
3.

The radical axis bisects common tangents of two circles.
4
.

Only concurrent circles do not have a radical axis.
5.

If two circles cut a third circle orthogonally, the radical axis
of the two circles will pass through the centre of the third
circle.
6.

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

The radical axis of three circles taken in pairs are concurrent at a
point called radical center.
Radical Center

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.
NOTE: Circle orthogonal to three circles

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).
1.
Family of circles

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.
2.
Family of circles

The equation of a family of circles passing through two
given points (x1, y1) & (x2, y2) can be written in the form:
3.
Family of circles

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,
4.
Family of circles

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
C. x2 + y2 - 3x - 9y = 0 D. x2 + y2 - 3x + 9y = 0
JEE Main 2015

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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