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Standard XII

Mathematics

Modulus of a Complex Number

lf 4, 4 and...

Question

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 $a - b + c =$

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Solution

The correct option is B $- 6$
$S \equiv ((x - 4)^{2} + (y - 4)^{2} = 0 \Rightarrow x^{2} + y^{2} - 8 x - 8 y + 32 = 0$
$S^{'} \equiv (x - 8)^{2} + (y - 2)^{2} = 0 \Rightarrow x^{2} + y^{2} - 16 x - 4 y + 68 = 0$
Hence radical axis of the given coaxial system is, $S - S^{'} = 0$
$\Rightarrow 8 x - 4 y - 36 = 0 \Rightarrow 2 x - y - 9 = 0$

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Similar questions

Q. lf

(1, 2)

is one limiting point and

x^{2} + y^{2}

-6x-

8 y + 25 = 0

is one circle of a coaxial system then the equation of the radical axis of the coaxial system is

Q. If

A (2, 1), B (0, 5)

are limiting points of a coaxial system then the radical axis of the conjugate coaxial system 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) =

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. If

x = 1

is the equation of the common radical axis and

(2, 3)

is one limiting point of a coaxial system, then the other limiting point is

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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 isax+by+c=0(a>0) . Then a−b+c=

The correct option is B −6S≡((x−4)2+(y−4)2=0⇒x2+y2−8x−8y+32=0S′≡(x−8)2+(y−2)2=0⇒x2+y2−16x−4y+68=0Hence radical axis of the given coaxial system is, S−S′=0⇒8x−4y−36=0⇒2x−y−9=0

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 $a - b + c =$