Find the
equation of the circle having the pair of lines x^{2} + 2xy+ 3x +6y =
0 as its normals and having the size just sufficient to contain the circle

x(x – 4) + y(y –3) = 0. .

Given circle is x (x –4) + y ( y –3) = 0

or x^{2} + y^{2} –4x
– 3y = 0 . . . (1).

Given pair of lines is

x (2) can be written as x \ 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 Band
its radius r_{1} =

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 + r_{1} = 5 +

Hence equation of required circle
will be (x+3)^{2} +

or x^{2} + y^{2} +
6x- 3y – 45 = 0

Note: If A lies insides circle
(1),then also radius of the required circle = AB + r_{1}. .

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+5c^{2}y=1
as c→ 1. .

Given lines are

(2 + c) x + 5c^{2}y = 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.

Show that
one of the circles x^{2} + y^{2} +2gx + c = 0 and x^{2}
+ y^{2} +2g¢x + c = 0 lies within the other, then
gg¢ and c are both positive. .

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¢| < .

&⇒
g^{2} + g¢^{2} – 2gg¢ < g^{2} – c + g¢^{2} – c – 2

&⇒
gg¢ - c >

&⇒
(gg¢)^{2} + c^{2} – 2cgg¢
> g^{2}g¢^{2} – c(g^{2} + g¢^{2}) + c^{2}

&⇒
c(g^{2} + g¢^{2} – 2gg¢) > 0

&⇒
c (g - g¢)^{2} > 0

&⇒ c > 0 ( since (g - g¢)^{2}
> 0 )

The line Ax
+ By + C = 0 cuts the circles x^{2} + y^{2} + ax + by + c = 0
at P and Q. The line A¢x + B¢y + C¢
= 0 cuts the circle x^{2} + y^{2} + a¢x
+ b¢y + c¢ = 0 at R and S. If P, Q, R and S are
concyclic, show that =
0.

The equation of circle through P and Q is

x^{2} + y^{2} + ax +
by + c + l (Ax + By + C) = 0 . .
. (1) .

and the equation of circle through R and S is

x^{2} + y^{2} + 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

Find the locus of the midpoints of chords of the circle

x^{2}
+ y^{2} + 2gx + 2fy + c = 0, which subtend a right angle at the
origin.

Let the mid point of
chord be P(h,k). Its equation will be T=S_{1} .

or x(h + g) + y(k + f)
– h^{2} – k^{2} - 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,

x^{2} + y^{2}
+ 2gx + 2fy + c+ l(x(h + g) + y(k + f) - h^{2}
- k^{2} - 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 ( -h^{2} – k^{2} – gh – fk) = 0

and -h = g + , - k = f +

&⇒
l
= -2

&⇒ required locus of (h, k) is;

2x^{2} + 2y^{2}
+ 2gh +2fy +c = 0. .