Question

The circles $x^2+y^2+ax=0$ & $x^2+y^2=c^2(c>0)$ touch each other if

• A. $a^2+c^2=1$
• B. $a^2-c^2=0$
• C. $a^2-c^2=1$
• D. None of these

Answer 1

The correct option is B $a^2-c^2=0$

Given circles:$c_1:x^2+y^2+ax=0$

$c_2:x^2+y^2=c^2$

General equation of circle:$x^2+y^2+2gx+2fy+c=0$

where $(-g,-f)$ is the co ordinate of center of the circle and $\sqrt{g^2+f^2-c}$ is the radius

$c_1:x^2+y^2+2x\left(\frac{a}{2}\right)=0$

$c_2:x^2+y^2=c^2$

$\therefore$Center of $c_1:(-\frac{a}{2},0)$

Center of $c_2:(0,0)$

Radius of $c_1:\sqrt{\frac{a^2}{4}+0-0}$

$=\frac{a}{2}$

Radius of $c_2:c$

(Image)

Condition of two circles to touch externally:

Sum of radius=distance between center

$\Rightarrow c+\frac{a}{2}=\sqrt{{\left(\frac{-a}{2}\right)}^2+0^2}$

$c+\frac{a}{2}=\frac{a}{2}$

$\Rightarrow c=0$

and if $c=0,c_2$ cannot exist

Now, if circles touch each other internally

Distance between centres =|difference in radius|

$\Rightarrow \frac{a}{2}=|c-\frac{a}{2}|$

If $c>\frac{a}{2}$

$\Rightarrow \frac{a}{2}=c-\frac{a}{2}$

$\Rightarrow a=c$

If $c<\frac{a}{2}$

$\frac{a}{2}=\frac{a}{2}-c\Rightarrow c=0$

Again $c_2$ won't exist if $c=0$

$\therefore$Condition for $c_1$ and $c_2$ to touch each other is $a=c$ (Internally)

$\Rightarrow a^2-c^2=0$