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Solved Subjective Questions on Circle Set 4

Posted on - 14-02-2017

JEE Math Circle

IIT JEE

Example 1.

Find the length of the chord of the circle x2 + y2 = 4 through which is of minimum length.

Solution

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)

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 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 .

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 pointto 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 - 6a. 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

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 = (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.

 
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