Mathematics - Circle Question with Solution | TestHub

Given,

$S_1:x^2+y^2-x-y-\frac{1}{2}=0$

So, the centre is $C_1:\left(\frac{1}{2}, \frac{1}{2}\right)$ and the radius is $r_1=\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{2}}=1$

$S_2:x^2+y^2-4y+\frac{7}{4}=0$

So, the centre is $C_2:\left(0,2\right)$ and the radius is 

$r_2=\sqrt{4-\frac{7}{4}}=\frac{3}{2}$

$S_3:x^2+y^2-4x-2y+5-r^2=0$

So, the centre is $C_3:\left(2,1\right)$ and the radius is 

$r_3=\sqrt{4+1-5+r^2}=\left|r\right|$

From the diagram, we can say that if $A\cup B\subseteq C,$ then the circle $A\hspace{0.17em}\&\hspace{0.17em}B$ should lie in the circle $C$

Here, $C_1{C}_3=\sqrt{\frac{5}{2}}$

So, $\left.\sqrt{\frac{5}{2}}\le \left|r-1\right|\Rightarrow \begin{array}{l}r\le 1-\sqrt{\frac{5}{2}}\\ r\ge 1+\sqrt{\frac{5}{2}}\end{array}\right\}$

Also, $C_2{C}_3=\sqrt{5}\le \left|r-\frac{3}{2}\right|$

$\Rightarrow \left.\begin{array}{l}r\ge \sqrt{5}+\frac{3}{2}\\ r\le \frac{3}{2}-\sqrt{5}\end{array}\right\}$

So, the minimum value of $r=\frac{3+2\sqrt{5}}{2}$