Question

Find the equation of the circle of minimum radius which contains the three circles
$S_1\equiv x^2+y^2-4y-5=0$
$S_2\equiv x^2+y^2+12x+4y+31=0$
and $S_2\equiv x^2+y^2+6x+12y+36=0$

Answer 1

The given circles are $S_1$ (0, 2), 3 $S_2$ (-6, - 2), 3 and $S_3$ (- 3, - 6), 3. In other words all the three circles are of same radius. Let the required circle be (h, k),r. The circle is to contain the three given circles and should be of minimum radius. Hence the given circles should touch the required circle internally so that the distance between the centres is equal to difference of radii, i.e., $r-3,r-3,r-3$ as all circles are of equal radius

$\therefore h^2+(k-2{)}^2=(h-6{)}^2+(k+2{)}^2=(h+3{)}^2+(k+6{)}^2=(r-3{)}^2$

$\therefore 6h+16k+41=0$ and $6h-8k-5=0$

Solving, we get (h, k) = $(-\frac{31}{18},-\frac{23}{12})$ and

$(r-3{)}^2$= $h^2+(k-2{)}^2={(-\frac{31}{18})}^2+{(-\frac{47}{12})}^2$

$=\frac{1}{(36{)}^2}\left(23725\right)=\frac{25\times 949}{(36{)}^2}$

$\therefore r-3=\frac{5}{36}\sqrt{949}$ or $r=3+\frac{5}{36}\sqrt{949}$

Hence,

the required circle is $(x-h{)}^2+(y-k{)}^2=r^2$