Question

Assertion :STATEMENT - 1 : If a line $L=0$ is tangent to the circle $S=0$, then it will also be a tangent to the circle $S+\lambda L=0$. Reason: STATEMENT - 2 : If a line touches a circle, then perpendicular distance of the line from the centre of the circle is equal to the radius of the circle.

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is A Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

$L=ax+by+c$ and $S\equiv x^2+y^2-r^2$
$\therefore r^2=\frac{c^2}{a^2+b^2}$
Now $S+\lambda L=x^2+y^2-r^2+\lambda ax+\lambda by+\lambda c$
Centre of the circle $S+\lambda L=0$ is $[-\frac{\lambda a}{2},-\frac{\lambda b}{2}]$
and radius of the circle $S+\lambda L=0$ is
$\sqrt{{\lambda }^2\frac{(a^2+b^2)}{4}-\lambda c+r^2}$
$\frac{\sqrt{\frac{{\lambda }^2}{4}(a^2+b^2)+c^2-\lambda c(a^2+b^2)}}{\sqrt{a^2+b^2}}$
$=\frac{\left|\frac{\lambda }{2}\right(a^2+b^2)-c|}{\sqrt{a^2+b^2}}$
Distance of $[-\frac{\lambda a}{2},-\frac{\lambda b}{2}]$ from
$L=0$ is $\frac{\left|\frac{\lambda }{2}\right(a^2+b^2)-c|}{\sqrt{a^2+b^2}}$
Therefore, $L=0$ is tangent to the circle $S+\lambda L=0$.
Thus STATEMENT-1 is true and is explained by STATEMENT-2.