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:

Consider the circles $S_1=x^2+y^2-2x-4y-4=0$ and $S_2=x^2+y^2+2x+4y+4=0$ & the line $L=x+2y+2=0$ then which of the following is not correct

• A. $L$ is the radical axis of $S_1$ & $S_2$
• B. $L$ is perpendicular to the line joining the centres $S_1$ & $S_2$
• C. $L$ is common chord of $S_1$ & $S_2$
• D. $L$ is common tangent of $S_1$ & $S_2$

Answer 1

Answer: (C), (D)

The correct options are
C $L$ is common chord of $S_1$ & $S_2$
D $L$ is common tangent of $S_1$ & $S_2$

Given $S_1=x^2+y^2-2x-4y-4=0$ and $S_2=x^2+y^2+2x+4y+4=0$

$\therefore$ Radical axis is $S_1-S_2=0$

$\Rightarrow x+2y+2=0$ which is line $L=0$ therefore the choice (A) is correct

Again for $S_1=0$ Centre is $(1,2)$ and radius $\sqrt{g^2+f^2-c}=3$

and $S_2=0$ gives Centre $(-1,-2)$ and radius 1.

Now slope of line joining the centres of $S_1=0$ & $S_2=0$ is

$\frac{y_2-y_1}{x_2-x_1}=\frac{2+2}{1+1}=2$ (say $m_1$) and the slope of the line $L=0$ is $\frac{coefficentofx}{coefficentofy}=-\frac{1}{2}$ (say $m_2$)

$\therefore m_1{m}_2=-1$

So line $L$ is $\perp$ to the line joining the centres of $S_1=0$ & $S_2=0$

$\therefore$ Choice (B) is correct

Again centre & radius of $S_1$ are $(1,2)$ & $3$ respectively

$\therefore$ Length of $\perp$er from $(1,2)$ to the line $x+2y+2=0$ is given by $\frac{|1+4+2|}{\sqrt{5}}=\frac{7}{\sqrt{5}}\ne radius$

$\therefore$ Line $L=0$ cannot be common tangent to $S_1$ & $S_2$

$\therefore$ Choice (D) is wrong

Also the circles $S_1=0$ & $S_2=0$ are not intersecting circles they have no common chord so choice (C) is wrong.

Hence choices C, D are not the correct statements.