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:

The equation of radical axis of the system of coaxial circles $x^2+y^2+2ax+2by+c+2\lambda (ax-by+1)=0$ is

• A. $bx+ay-1=0$
• B. $2(ax+by)-1=0$
• C. $2(bx-ay)+1=0$
• D. $ax-by+1=0$

Answer 1

The correct option is D $ax-by+1=0$

For any value of $\lambda$ the given equation

$(x^2+y^2+2ax+2by+c)+2\lambda (ax-by+1)=0$

represents the circle passing through the intersection of the circle

$x^2+y^2+2ax+2by+c=0$ & the line
$ax-by+1=0$

So the straight line $ax-by+1=0$ is the common chord
for each circle of the system

Hence choice (d) is the correct answer.