Example
1.
Find the length of the chord of the
circle x2 + y2 = 4 through 
which
is of minimum length.
Solution
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1. Given
circle is x2 + y2 = 22 .
. . (1).
Its
centre C º (0, 0)
Let
P º .
For
point P, x2 + y2 –4
= 1+  - 4 = - < 0 ,
Hence
point P lies inside the circles (1)
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Let
AB be any chord of circle (1) through P 
. Let CL ^AB,
then L will be the mid-point of AB.
Now
AB = 2AL = 
=
= 
. . . (2)
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since
C and P are fixed points , therefore CP is fixed . .
from
(2) , AB will be minimum if LP2 is minimum and minimum value of
LP2 = 0, when P coincides with L. .
Hence
for minimum length of AB, CP ^AB and P is the mid-point of AB.
Now
from (2) minimum value of AB
=
2 =  units .
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Example
2.
Show that all the
chords of the curve 3x2 - y2 - 2x + 4y = 0 which subtend
a right angle at the origin are concurrent. Does this result hold for the curve
3x2 + 3y2 - 2x + 4y = 0 ? If yes, what is the point of
concurrency and if not, give reasons.
Solution
Let the
chord be lx + my = l and it subtends a right angle at the origin. Making the
curve 3x2 – y2 – 2x + 4y = 0 homogeneous by the help of
lx + my = 1, we get.
3x2
– y2 – 2x (lx + my) + 4y (lx + my) = 0
Since the
lines are perpendicular each other,
3 – 2l + 4m
– 1 = 0 or –2l + 4m + 2 = 0
&⇒ l – 2m = 1
Hence, the
chord passes through the point (1, –2)
Now, if the
curve be a circle, then 3x2 + 3y2 – 2x + 4y = 0
&⇒ 
,
whose centre is 
.
All the chords of this circle which subtend a right angle at the origin, which
lies on the circumference, must be diameters. Hence they pass through the
centre .
Example
3.
(a). Let a circle
be given by 2x(x - a) + y(2y - b) = 0, a > 0, b > 0. Find the
condition on a and b, if two chords, each bisected by the x axis, can be drawn
to the circle from the point (a, b/2).
(b).Find
the intervals of values of a for which the line y + x = 0 bisects two chords
drawn from a point
to
the circle 2x2+2y2-(1+Ö2a)x- (1-Ö2a)y=
0.
Solution
(a). The
given circle is 2x (x – a) + y (2y –b) = 0.
&⇒ 2x2 – 2ax + 2y2
– by = 0
&⇒ x2 + y2 – ax
– 
y
= 0
Equation of
any chord of the circle whose mid-point is (h, o)
hx + o – 
(x
+ h) – 
(y
+ 0) = h2 + 0 – ah
&⇒ 
Since it
passes through 
;
&⇒ ah – 
&⇒ 
&⇒ 8h2 – 12ah + 4a2
+ b2 = 0
Since roots
of this equation should be real and distinct,
144a2
– 4 ´ 8 (4a2 + b2) > 0
&⇒ 9a2 – 8a2 – 2b2
> 0
&⇒ a2 > 2b2
(b). Let y+ x = 0
bisects a chord through 
at
(h, k). Then h + k = 0.
Also (h, k) being
mid-point of this chord of circle
x2 + y2
- 
x
- 
y
= 0
Its equation is hx + ky
- (x
+ h) - (y
+ k)
= h2 + k2
- h
- k
&⇒
This passes through P
so
i.e. 8h2 -
6
a.
h + 1 + 2a2 = 0
h should be real and
distinct for having two such chords. .
\ Discriminant > 0. so a2
> 2 \ a ∈ (-¥, -2) È
(2, ¥).
Example
4.
Find the radius of the
smallest circle which touches the line 3x - y = 6 at (1, -3) and also touches
the line y = x. Compute the radius approximately.
Solution
The circle is (x -1)2
+ (y + 3)2 + l(3x – y – 6) = 0 ….(6) .
which touches line 3x –
y = 6 at (1, -3)
It can be written as x2
+ y2 + (3l - 2)x + (6 - l)y
+ (10 - 6l) = 0 …(2)
This touches y = x
\
Radius = length of perpendicular from centre 
on
x – y = 0
i.e. 
&⇒
l
= - 8 + 4
\
(Radius)2 = 
&⇒
Radius = 
(-
8 + 4)
= 
.
Example
5.
Show that the common
tangents to the circles x2 + y2 – 6x=0 and x2+
y2+2x= 0 form an equilateral triangle.
Solution
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Centre
of circles are (3, 0) and (-1, 0). So they lie on x-axis .
Distance
between centres = 4
Sum
of radii = 3 + 1 = 4. So both touch each other externally at origin. .
Let
A(-h, 0) be the point where the tangents meet. .
A
divides PQ in ratio 3 : 1
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i.e. AQ : AP = 3 : 1 .
\
- h = 
&⇒
h = 
\
A (-3, 0). .
The pair of tangents
from A is (-3x + x – 3)2 = (x2 + y2 + 2x) ( 9
– 6)
&⇒
x2 – 3y2 + 6x + 9 = 0 …(1)
The common tangent BC
is x = 0 ….(2) .
(1) and (2) intersect
at y = ± 
\
B(0, ),
C(0, -).
AC = AB = 
and
BC = 2
\
DABC
is equilateral.
