The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles…
The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles $x^2+y^2-4 x-4 y+6=0$ and $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+\lambda=0$ is the interval:
$(12,32)$
$(18,42)$
$(12,24)$
$(18,48)$
Solution
The equations of the circles are
$
x^2+y^2-10 x-10 y+\lambda=0
$
and $x^2+y^2-4 x-4 y+6=0$
$
\begin{aligned}
&C_1=\text { centre of }(1)=(5,5) \\
&\mathrm{C}_2=\text { centre of }(2)=(2,2) \\
&d=\text { distance between centres } \\
&=\mathrm{C}_1 \mathrm{C}_2=\sqrt{9+9}=\sqrt{18} \\
&r_1=\sqrt{50-\lambda}, r_2=\sqrt{2} \\
&\text { For exactly two common tangents we } \\
&\text { have } \\
&r_1-r_2 < C_1 C_2 < r_1+r_2 \\
&\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\
&\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} \text { or } 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\
&\Rightarrow \sqrt{50-\lambda} < 4 \sqrt{2} \text { or } 2 \sqrt{2} < \sqrt{50-\lambda} \\
&\Rightarrow 50-\lambda < 32 \text { or } 8 < 50-\lambda \\
&\Rightarrow \lambda>18 \text { or } \lambda < 42 \\
&\text { Required interval is }(18,42) \\
&
\end{aligned}
$