For a coaxial system x2-y2-2- y-c-0 radical axis is

The correct option is B y=0 x^2+y^2+2μ y+c=0 is a coaxial system of circle nbsp;So, s_1:x^2+y^2+2μ _1y+c=0 s_2:x^2+y^2+2μ _2y+c=0 So, ...

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Standard XII

Mathematics

Slope Point Form of a Line

For a coaxial...

Question

For a coaxial system $x^{2} + y^{2} + 2 μ y + c = 0$ radical axis is

x = 0

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y = 0

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y = c

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y + c = 0

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Solution

The correct option is B $y = 0$
$x^{2} + y^{2} + 2 μ y + c = 0$ is a coaxial system of circle So, $s_{1} : x^{2} + y^{2} + 2 μ_{1} y + c = 0$
$s_{2} : x^{2} + y^{2} + 2 μ_{2} y + c = 0$
So, $e q^{n}$ of radical axis of system is
$s_{1} - s_{2} = 0$
$y (μ_{1} - μ_{2}) = 0$
$\Rightarrow y = 0$

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Similar questions

Q. Assertion (A): The conjugate co-axial system of

x^{2} + y^{2} + 2 λ x + c = 0

x^{2} + y^{2} + 2 μ y - c = 0

Reason (R): the radical axis of given system is perpendicular to the line of centres of conjugate system

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:

The equation of radical axis of the system of coaxial circles

x^{2} + y^{2} + 2 a x + 2 b y + c + 2 λ (a x - b y + 1) = 0

Q. If ax

+

+ c a = 0

, where a,b,c>0 is the equation of the common radical axis of the coaxial system

(3 x^{2} + 3 y^{2} - 2 x - 2 y - 1) + λ (x^{2} + y^{2} - x - 2 y - 3) = 0

, then

a b + b c +

=

Q. lf

(4, 4)

and

(8, 2)

are limiting points of a coaxial system then the equation of the common radical axis of the coaxial system is

a x + b y + c = 0 (a > 0)

. Then

a - b + c =

Q. The equation of the common radical axis of the orthogonal coaxial system of the coaxial system

(x^{2} + y^{2} - 4 x - 6 y + 5) + λ (2 x + 3 y + 4) = 0

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For a coaxial system x2+y2+2μy+c=0 radical axis is

The correct option is B y=0x2+y2+2μy+c=0 is a coaxial system of circle So, s1:x2+y2+2μ1y+c=0s2:x2+y2+2μ2y+c=0So, eqn of radical axis of system iss1−s2=0y(μ1−μ2)=0⇒y=0

For a coaxial system $x^{2} + y^{2} + 2 μ y + c = 0$ radical axis is

The correct option is B $y = 0$
$x^{2} + y^{2} + 2 μ y + c = 0$ is a coaxial system of circle So, $s_{1} : x^{2} + y^{2} + 2 μ_{1} y + c = 0$
$s_{2} : x^{2} + y^{2} + 2 μ_{2} y + c = 0$
So, $e q^{n}$ of radical axis of system is
$s_{1} - s_{2} = 0$
$y (μ_{1} - μ_{2}) = 0$
$\Rightarrow y = 0$