Example 1.
Find the
equation of the circle having the pair of lines x2 + 2xy+ 3x +6y =
0 as its normals and having the size just sufficient to contain the circle
x(x – 4) + y(y –3) = 0. .
Solution
Given circle is x (x –4) + y ( y –3)
= 0
or x2 + y2 –4x
– 3y = 0 . . . (1).
Given pair of lines is
|
x2 + 2xy +3x + 6y = 0 .
. . (2).
(2) can be written as
x2 + (2y +3) x + 6y = 0
\
x = 
or 2x + 2y +3 = ±
\
2x + 6 = 0 i.e. x +3 = 0 . . . (3).
and 2x + 4y = 0 or x + 2y = 0 .
. . (4).
|
|
Thus (3) and (4) are the lines
represented by equation(2). According to question lines (3) and (4) are the
two normals to the circle whose equation is to be found. .
Solving (3) and (4), we get x = -3,
y = 3/2
\
Centre of the required circle will be A
Centre of circle (1) is B
and
its radius r1 = 
AB = 
\
A lies outside circle (1)
Since the required circle should
have size just sufficient to contain circle (1)
\ radius of the
required circle = AB + r1 = 5 + 
Hence equation of required circle
will be (x+3)2 + 
or x2 + y2 +
6x- 3y – 45 = 0
Note: If A lies insides circle
(1),then also radius of the required circle = AB + r1. .
Example 2.
Find the
equation of the circle which passes through (2, 0) and whose centre is the
limit of the point of intersection of the lines 3x +5y = 1 and (2+c)x+5c2y=1
as c→ 1. .
Solution
Given lines are
(2 + c) x + 5c2y = 1 .
. . . (1).
3x + 5y = 1 . .
. . (2).
Solving (1) and (2), we get
point of intersection when c →
1, will be given by
i.e. 
is the center of the circle
Therefore, the equation of the
required circle is
i.e. ( 25x – 10)2
+ (25y +1)2 = 1601.
Example 3.
Show that
one of the circles x2 + y2 +2gx + c = 0 and x2
+ y2 +2g¢x + c = 0 lies within the other, then
gg¢ and c are both positive. .
Solution
The radical axis of the given circles
is 2(g - g¢)x = 0
i.e. x = 0 ( y-axis ) .
Since one lies within
other so they neither intersect nor touch. So radical axis neither intersects
them nor touch them. Hence both circles lie on the same side of the radical
axis (y-axis). .
So the x-coordinate of their centres
are of same sign. Hence gg¢ > 0. .
Next, distance between
centres < difference of radii and the centres are (-g, 0),
(-g¢, 0), so |-g + g¢| < 
.
&⇒
g2 + g¢2 – 2gg¢ < g2 – c + g¢2 – c – 2
&⇒
gg¢ - c >
&⇒
(gg¢)2 + c2 – 2cgg¢
> g2g¢2 – c(g2 + g¢2) + c2
&⇒
c(g2 + g¢2 – 2gg¢) > 0
&⇒
c (g - g¢)2 > 0
&⇒ c > 0 ( since (g - g¢)2
> 0 )
Example 4.
The line Ax
+ By + C = 0 cuts the circles x2 + y2 + ax + by + c = 0
at P and Q. The line A¢x + B¢y + C¢
= 0 cuts the circle x2 + y2 + a¢x
+ b¢y + c¢ = 0 at R and S. If P, Q, R and S are
concyclic, show that 
=
0.
Solution
The equation of circle through P and
Q is
x2 + y2 + ax +
by + c + l (Ax + By + C) = 0 . .
. (1) .
and the equation of circle through R
and S is
x2 + y2 + a¢x
+ b¢y + c¢ + m (A¢x
+ B¢y + C¢) = 0 . . . . (2) .
If P, Q, R and S are concyclic, then
(1) and (2) represent the same circle. .
&⇒ a
+ lA = a¢+ mA¢
b + lB = b¢
+ mB¢
c + lC = c¢
+ mC¢
&⇒ a - a¢
+ lA - mA¢ = 0
b - b¢ + lB
- mB¢ = 0
c - c¢ + lC
- mC¢ = 0
Eliminating l
and m, we get
= 0 or 
= 0
Example 5.
Find the locus of the
midpoints of chords of the circle
x2
+ y2 + 2gx + 2fy + c = 0, which subtend a right angle at the
origin.
Solution
Let the mid point of
chord be P(h,k). Its equation will be T=S1 .
or x(h + g) + y(k + f)
– h2 – k2 - gh –fk = 0
Now equation of any
circle that can be drawn to pass through the points where this chord
and given circle meet, will be,
x2 + y2
+ 2gx + 2fy + c+ l(x(h + g) + y(k + f) - h2
- k2 - gh - fk) = 0
Since the chord
subtends a right angle at the origin, this circle must pass through (0, 0)
and have it’s centre at (h, k)
&⇒
c + l ( -h2 – k2 – gh – fk) = 0
and -h = g + 
,
- k = f + 
&⇒
l
= -2
&⇒ required locus of (h, k) is;
2x2 + 2y2
+ 2gh +2fy +c = 0. .
