Question

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 $\perp$ 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 $S_1+\lambda S_2=0,\lambda$ 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 $x^2+y^2+2gx+c=0$ where $g$ is variable & $c$ is constant

If $g=\pm \sqrt{c}$ 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 $x^2+y^2+2gx+c=0$

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 $(-1,2)$ & $(3,4)$ is

• A. $x^2+y^2+4x+3y=0$
• B. $x^2+y^2-4x+3y=0$
• C. $x^2+y^2+4x-3y=0$
• D. $x^2+y^2-4x-3y=0$

Answer 1

The correct option is C $x^2+y^2+4x-3y=0$

As limiting points are $(-1,2)$ & $(3,4)$

$\therefore$ Point circles of the coaxial system are given by

${(x+1)}^2+{(y-2)}^2=0$ and

${(x-3)}^2+{(y-4)}^2=0$

or $x^2+y^2+2x-4y+5=0$ & $x^2+y^2+6x-8y+25=0$

$\therefore$ Equation of coaxial system is given by

$(x^2+y^2+2x-4y+5)+\lambda (x^2+y^2-6x-8y+25)=0$ .....(1)

Now equation (1) passess through $(0,0)$

So ,$\lambda =-\frac{1}{5}$

$\therefore$ From equation (1) we have

$5(x^2+y^2+2x-4y+5)-(x^2+y^2-6x-8y+25)=0$

$5(x^2+y^2+2x-4y+5)-(x^2+y^2-6x-8y+25)=0$

$x^2+y^2+4x-3y=0$

Hence choice (C) is correct answer.