ANAND CLASSES-Best NDA Coaching Center In Jalandhar Punjab-Coaching Classes For NDA in Punjab

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Call – 9463138669 – ANAND CLASSES
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A circle is a set of points in the
plane which is equidistant from a
fixed point.
DEFINITION OF A CIRCLE
The fixed point is called as centre.
.
P
C
r
The distance between the centre and any point on the
Circle is called Radius.

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 If the radius is one unit, then the circle is called a
unit circle.
 If the radius is zero, then the circle is called a point
circle.

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EQUATION OF A CIRCLE
 The equation of a circle with centre at the origin and
radius r is x2+y2 = r2
Theorem: The equation of the circle with centre
C (h,k) and radius r is (x-h)2+(y-k)2=r2
Proof : Let P(x1,y1) be a point on the locus
P lies on the circle
 x1−h 𝟐+ y1−k 𝟐 = r
 The locus of P is (x-h)2+(y-k)2 = r2
 CP=r
 (x1-h)2 + (y1-k)2 = r2
C(h,k)
P(x1, y1)
r

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Theorem: If g2+f2-c0 then the equation x2+y2+2gx+2fy+c=0
represents a circle with centre (-g,-f ) and radius is g𝟐 + f𝟐
−c
Proof :
Given equation is x2+y2+2gx+2fy+c=0
 x2+2gx+g2+y2+2fy+f2 = g2+f2-c
 (x+g)2+(y+f)2 = g2+f2-c
 x2+2gx+y2+2fy =-c
adding g2+f2 on both sides.

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It represents a circle
Centre = (-g, -f)
 [x-(-g)]2+[y-(-f)]2 = g𝟐 + f𝟐
−c
𝟐
Radius = r = g𝟐 + f𝟐
−c

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 If g2+f2-c>0, equation(i) represents a real circle with
centre (-g,-f ).
 If g2+f2-c=0, the equation(i) represents a circle
whose centre is (-g,-f) and radius is zero. i.e. the
circle coincides with the centre and so it
represents a point (-g,-f) called as a point circle.
NOTE
[x-(-g)]2+[y-(-f)]2 = g𝟐 + f𝟐
−c
𝟐
...............(i)

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 If g2+f2-c<0 radius of the circle is imaginary. In this
case, there are no real points on the circle and so it is
called a virtual circle or imaginary circle.

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Theorem: The conditions that the equation
ax2+2hxy+by2+2gx+2fy+c = 0 to represent a circle are
i) a = b  0 ii) h = 0 iii) g2+f2-ac0
Proof :
We know that equation of a circle is of the form
x2+y2+2Gx+2Fy+C=0 (1)
If ax2+2hxy+by2+2gx+2fy+c=0 (2)
represents a circle
Comparing (1) and (2) we get
a
1
=
b
1
=
g
G
=
f
F
=
c
C
and h = 0

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 a = b, h = 0
G2+F2-C  0
⇒
g
a
𝟐
+
f
a
𝟐
−
c
a
≥ 0
 g2+f2-ac  0

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 In every equation of a circle
coefficient of x2 = coefficient of y2 and
coefficient of xy =0
 If ax2+ay2+2gx+2fy+c = 0 represents a circle, then its
centre =
−g
a
,
−f
a
and its
radius =
g𝟐
+f
𝟐
−c
|a|

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 If the circle S=0 and the line L=0
intersect, then the equation of the
circle passing through the points of
intersection of the circle and the line
is S+L=0 where  is a parameter
S=0
L=0
 If a line passing through a point P(x1,y1) intersects the
circle S=0 at the points A and B then PA.PB=S11 .
P
A B

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DEFINITIONS
Let A, B be two points on the circle, then
1. A secant is a line that intersects the circle at two points.
SECANT LINE
B
A

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A B
Chord on
the circle
AB=length of chord
A B
C
P
2. A segment of line joining the
points ‘A’ and ‘B’ is called chord.
3. AB is called the
AB of the circle.

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Theorem: The angle in a semicircle is
a right angle.
Let O=(0,0) be the centre and
A(r,0), B(-r,0) be the ends of the
diameter AB of the circle
x2+y2=r2.
Let P(x1,y1) be a point on the circle.
Proof :
 x1
2 + y1
2 = r2
x'
B(-r,0)
x
A(r,0)
P(x1,y1)
y
y'
O
Equation of a
circle with centre
(0,0) and radius r
is x2+y2=r2

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 Slope of PA is (m1) =
y𝟏
−0
x𝟏−r
 Slope of PB is (m2) =
y𝟏
−0
x𝟏+r
 m1m2 =
y𝟏
x𝟏−r
.
y𝟏
x𝟏+r =
y𝟏
𝟐
x𝟏
𝟐
−r𝟐 =
y𝟏
𝟐
−y𝟏
𝟐
= -1
(∵ x1
2 – r2 = -y1
2)
 m1m2 = -1  PA to PB
 APB = 900

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Theorem:- The equation of a circle
having the segment joining A(x1,y1),
B(x2,y2) as diameter is
(x-x1) (x-x2)+(y-y1)(y-y2) = 0
Proof :
Given that A(x1,y1), B(x2,y2) are the
ends of the diameter
APB = 900
Slope of PA  Slope of PB = -1
Let P(x, y) be any point on the circle.
 
,
P x y
A
(x1,y1)
B
(x2,y2)

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 (y-y1) (y-y2) = - (x-x1) (x-x2)
 (x-x1) (x-x2) + (y-y1) (y-y2)=0
Equation of a circle having AB as diameter is
x2+y2-(x1+x2)x-(y1+y2)y+x1x2+y1y2 = 0
y − y𝟏
x − x𝟏
y − y𝟐
x − x𝟐
= −1

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1. Centre of the circle x2+y2+2gx+2fy+c=0 is…
1) (g,f)
2) (-g,f)
3) (-g,-f)
4) (g,-f)
KEY : 3

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2. Radius of the circle x2+y2+2gx+2fy+c=0 is…
1) g𝟐 + f𝟐
+c
2) g𝟐 + f𝟐
−c
3) g𝟐 − f𝟐
−c
4) g𝟐 − f𝟐
+c
KEY : 2

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3. Centre of the circle ax2+ay2+2gx+2fy+c=0 is…
1)
g
a
,
f
a
2)
−g
a
,
f
a
3)
−g
a
,
−f
a
4)
g
a
,
−f
a
KEY : 3

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4. Centre of the circle x2+y2+4x+6y-12=0 is…
1) (2,3)
2)(−2,−3)
3)(4,6)
4)(−4,−6)
KEY : 2

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5. Radius of the circle x2+y2+4x+6y-12=0 is…
1) 25
2) 5
3) 1
4) 2
KEY : 2

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S = x2+y2+2gx+2fy+c
NOTATIONS
S1 = xx1+yy1+g(x+x1)+f(y+y1)+c
S11 = x1
2+ y1
2+ 2gx1+2fy1+c
S12 = x1x2+y1y2+g(x1+x2)+f(y1+y2)+c

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POWER OF A POINT
Suppose S=0 is the equation of circle with centre ‘C’. Let
P(x1,y1) be any point in the plane then CP2-r2 is defined
as power of ‘p’ w.r.t to S = 0
i) If ‘P’ exterior of circle then power is positive
ii) If ‘P’ on the circle then power is zero
ii) If ‘P’ interior of circle then power is negative.

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POSITION OF A POINT
Let S = 0 be a circle in a plane and
P(x1,y1) be any point in the same plane
Then
(i) ‘P’ is the interior of circle
⇔ S11 < 0
Since P lies inside the circle, CP<r
⇔ CP2 < r2
⇔(x1+g)2+(y1+f)2 < r2
c
p

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⇔ x1
2+ y1
2+2gx1+2fy1+g2+f2 < g2+f2-C
⇔ x1
2+ y1
2+2gx1+2fy1+c < 0
⇔ S11 < 0
ii) P lies on the circle
⇔cp2-r2= 0
⇔ S11 = 0

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iii) P lies outside the circle
⇔cp2-r2> 0
⇔ S11 > 0

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CONCYCLIC POINTS
Theorem: If the lines a1x+b1y+c1 = 0, a2x+b2y+c2 = 0
cuts the coordinate axes in concyclic points then
prove that a1a2 = b1b2
Proof :
Given that L1  a1x+b1y+c1 = 0
L2  a2x+b2y+c2 = 0

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Meets the coordinate axes at ‘P’ and
‘Q’, L2=0 meets at R and S
P
Q
R
S
O X
Y
Then coordinates of P,Q,R,S are
P
−c𝟏
a
, 0 ; Q 0,
−c𝟏
b𝟏
;
R
−c𝟐
a𝟐
, 0 ; S 0,
−c𝟐
b𝟐

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Since P,Q,R,S are concyclic OP.OR = OQ.OS
−c𝟏
a𝟏
−c𝟐
a𝟐
=
−c𝟏
b𝟏
−c𝟐
b𝟐
 a1a2 = b1b2

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 If the lines a1x+b1y+c1 = 0, a2x+b2y+c2 = 0, meet the
coordinate axes in four distinct concyclic points, then
the equation of the circle passing through these
concyclic points is
(a1x+b1y+c1) (a2x+b2y+c2)- (a1b2+a2b1)xy = 0

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Theorem: The equation of the
circumcircle of the triangle formed by
the line ax+by+c= 0 with coordinate
axes is ab(x2+y2)+c(bx+ay) = 0
Proof :
The line ax+by+c=0 cuts
the coordinate axes at A
−c
a
, 0 , B 0,
−c
b
A
B
O
X
Y

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 AB is a diameter of circle, so the equation of circle is
x+
c
a
(x−0)+(y−0) y+
c
b
= 0
 x2+y2+c
x
a
+
y
b
= 0
 ab(x2+y2)+c(bx+ay) = 0
(x−x1)(x−x2)+(y−y1)(y−y2)=0

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CONCENTRIC CIRCLES
 Two circles are said to be concentric if their centres
are same
 The equation of any circle concentric with the circle
x2+y2+2gx+2fy+c=0 is x2+y2+2gx+2fy+=0, where  is
any real constant.

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1. The position of the point (1,2) with
respect to the circle x2+y2+6x+8y-96=0
is…
1) outside the circle
2) inside the circle
3) on the circle
4) none
KEY : 2

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2. The power of the point (2,4) with respect to the circle
x2+y2-4x-6y-12=0 is…
1) -48
2) 48
3) 24
4) -24
KEY : 4

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