The number of real circles belonging to the coaxial system x2-y2-2-x-c-0- is a variable and whose centre lie between the limiting points of the given coaxial system is

The correct option is C 0 (x+λ)^2+y^2+c λ^2=0 For λ^2=c The two point circle (x±√(c))^2+y^2=0 are called limiting points.No real circle of ...

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Standard XII

Mathematics

Position of a Point with Respect to Circle

The number of...

Question

The number of real circles belonging to the coaxial system $x^{2} + y^{2} + 2 λ x + c = 0 (λ$ is a variable) and whose centre lie between the limiting points of the given coaxial system is

1

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0

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Solution

The correct option is C $0$
$(x + λ)^{2} + y^{2} + c - λ^{2} = 0$
For $λ^{2} = c$
The two point circle $(x \pm \sqrt{c})^{2} + y^{2} = 0$ are called limiting points.
No real circle of this coaxial system have their centre between the limiting point.

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Similar questions

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

x^{2} + y^{2} + 2 λ x - 5 = 0

are

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) =

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:

Equation of circle through the origin & belonging to the coaxial system, of which the limit points are

(- 1, 2)

(3, 4)

Q. The equation of the circle of least radius belonging to the coaxial system of circles
orthogonal to the system

x^{2} + y^{2} + 2 λ x + 4 = 0

Q. 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

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The number of real circles belonging to the coaxial system x2+y2+2λx+c=0(λ is a variable) and whose centre lie between the limiting points of the given coaxial system is

The correct option is C 0(x+λ)2+y2+c−λ2=0For λ2=cThe two point circle (x±√c)2+y2=0 are called limiting points.No real circle of this coaxial system have their centre between the limiting point.

The number of real circles belonging to the coaxial system $x^{2} + y^{2} + 2 λ x + c = 0 (λ$ is a variable) and whose centre lie between the limiting points of the given coaxial system is

The correct option is C $0$
$(x + λ)^{2} + y^{2} + c - λ^{2} = 0$
For $λ^{2} = c$
The two point circle $(x \pm \sqrt{c})^{2} + y^{2} = 0$ are called limiting points.
No real circle of this coaxial system have their centre between the limiting point.