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Solved Subjective Questions on Circle Set 9

Posted on - 26-02-2017

JEE Math Circle

IIT JEE

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 Band 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. .

 
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