Find the length of the chord of the
circle x^{2} + y^{2} = 4 through which
is of minimum length.

1. Given
circle is x Its centre C º (0, 0) Let P º . For
point P, x Hence point P lies inside the circles (1) |

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)

since C and P are fixed points , therefore CP is fixed . . from
(2) , AB will be minimum if LP 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 . |

Show that all the
chords of the curve 3x^{2} - y^{2} - 2x + 4y = 0 which subtend
a right angle at the origin are concurrent. Does this result hold for the curve

3x^{2} + 3y^{2} - 2x + 4y = 0 ? If yes, what is the point of
concurrency and if not, give reasons.

Let the
chord be lx + my = l and it subtends a right angle at the origin. Making the
curve 3x^{2} – y^{2} – 2x + 4y = 0 homogeneous by the help of
lx + my = 1, we get.

3x^{2}
– y^{2} – 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 3x^{2} + 3y^{2} – 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 .

(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 pointto
the circle 2x^{2}+2y^{2}-(1+Ö2a)x- (1-Ö2a)y=
0.

(a). The given circle is 2x (x – a) + y (2y –b) = 0.

&⇒ 2x^{2} – 2ax + 2y^{2}
– by = 0

&⇒ x^{2} + y^{2} – ax
– y
= 0

Equation of any chord of the circle whose mid-point is (h, o)

hx + o – (x
+ h) – (y
+ 0) = h^{2} + 0 – ah

&⇒

Since it passes through ;

&⇒ ah –

&⇒

&⇒ 8h^{2} – 12ah + 4a^{2}
+ b^{2} = 0

Since roots of this equation should be real and distinct,

144a^{2}
– 4 ´ 8 (4a^{2} + b^{2}) > 0

&⇒ 9a^{2} – 8a^{2} – 2b^{2}
> 0

&⇒ a^{2} > 2b^{2}

(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

x^{2} + y^{2}
- x
- y
= 0

Its equation is hx + ky - (x + h) - (y + k)

= h^{2} + k^{2}
- h
- k

&⇒

This passes through P so

i.e. 8h^{2} -
6a.
h + 1 + 2a^{2} = 0

h should be real and distinct for having two such chords. .

\ Discriminant > 0. so a^{2}
> 2 \ a ∈ (-¥, -2) È
(2, ¥).

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.

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

Show that the common
tangents to the circles x^{2} + y^{2} – 6x=0 and x^{2}+
y^{2}+2x= 0 form an equilateral triangle.

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 |

i.e. AQ : AP = 3 : 1 .

\
- h =

&⇒
h =

\ A (-3, 0). .

The pair of tangents
from A is (-3x + x – 3)^{2} = (x^{2} + y^{2} + 2x) ( 9
– 6)

&⇒
x^{2} – 3y^{2} + 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.