The condition for the coaxial system x2-y2-2- x-c-0- where - is a parameter and c is a constant- to have distinct limiting points is

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Standard XII

Physics

A Non-Uniform Rod

The condition...

Question

The condition for the coaxial system $x^{2} + y^{2} + 2 λ x + c = 0$ , where $λ$ is a parameter and $c$ is a constant, to have distinct limiting points is

c = 0

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c < 0

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c = - 1

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c > 0

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Solution

The correct option is B $c > 0$
Radical axis is $x = 0$
System have distinct limiting points.
So, $r < d$
$d$ is the distance of centre $(λ, 0)$ from line $x = 0$
So,
$d = \frac{λ}{1} = λ$
$∴ r < d$
$λ^{2} - c < λ^{2}$
$c > 0$

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Similar questions

P (- 2, - 1)

and

0 (0, 3)

are the limiting points of a coaxial system of which

C = x^{2} + y^{2} + 5 x + y + 4 = 0

is a member. The circle

S = x^{2} + y^{2} - 4 x - 15 = 0

is orthogonal to the circle

C = 0

. The point where the polar of

P

with respect to

C = 0

cuts the circle

S = 0

A system of circles is said to be coaxial when every pair of circles has the same radical axis. For coaxial circles, we note that

(1)The centers of all coaxial circles lie in a straight line, which is

⊥

to the common radical axis

(2)Circles passing through two fixed points form a coaxial system with a line joining the points as a common radical axis.

(3)The equation to a coaxial system whose two members are

S_{1} = 0

S_{2} = 0

is given by

S_{1} + λ S_{2} = 0, λ

is parameter.

If we take line of centres as x -axis & common radical axis as y-axis, then the simplest form of equation of coaxial circles is given by

x^{2} + y^{2} + 2 g x + c = 0

where

g

is variable &

c

is constant

g = \pm \sqrt{c}

then radious

g^{2} - c

vanishes & the circle become a point circle. The points

(\pm \sqrt{c}, 0)

are called the limiting points of the system of coaxial circle given by

x^{2} + y^{2} + 2 g x + c = 0

On the basis of above information answer the following question:

(2, 1)

is a limit point of a coaxial system of circles containing

x^{2} + y^{2} - 6 x - 4 y - 3 = 0

then other limit point is

Q. lf

(1, 2)

and

(3, 4)

are limiting points of the given coaxial system then the least circle belonging to the orthogonal coaxial system is

x^{2} + y^{2} + a x + b y + c = 0

. Then

(a, c) =

Q. The differential equation of family of coaxial circle

x^{2} + y^{2} + 2 λ x + c = 0

Q. The number of limiting points of orthogonal coaxial system of the coaxial system

x^{2} + y^{2} + 2 λ x - 5 = 0

are

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The condition for the coaxial system x2+y2+2λx+c=0, where λ is a parameter and c is a constant, to have distinct limiting points is

The correct option is B c>0Radical axis is x=0System have distinct limiting points.So, r<dd is the distance of centre (λ,0) from line x=0So,d=λ1=λ∴r<dλ2−c<λ2c>0

The condition for the coaxial system $x^{2} + y^{2} + 2 λ x + c = 0$ , where $λ$ is a parameter and $c$ is a constant, to have distinct limiting points is

The correct option is B $c > 0$
Radical axis is $x = 0$
System have distinct limiting points.
So, $r < d$
$d$ is the distance of centre $(λ, 0)$ from line $x = 0$
So,
$d = \frac{λ}{1} = λ$
$∴ r < d$
$λ^{2} - c < λ^{2}$
$c > 0$