[[1]] is the equation of the circle of minimum radius which contains the three circles.

$S 1 \equiv x^{2} + y^{2} - 4 y - 5 = 0$

$S 2 \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0$

$S 3 \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0 .$

Question

[[1]] is the equation of the circle of minimum radius which contains the three circles.S1≡x2+y2-4y-5=0S2≡x2+y2+12x+4y+31=0S3≡x2+y2+6x+12y+36=0.

Accepted Answer

The general form of the equation of the circle

x^{2} + y^{2} + 2 gx + 2 fy + c = 0

, where the center of the circle is (−g,−f), and the radius is

\sqrt{g^{2} + f^{2} - c}

⇒For the equation

S 1 \equiv x^{2} + y^{2} - 4 y - 5 = 0

of the circle, we get to the center,

C 1 = (0, 2)

radius r1=3
⇒ For the equation

S 2 \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0

of the circle, we get to the center,

C 2 = (- 6, - 2)

and radius r2=3
⇒ For the equation

S 3 \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0

of the circle, we get to the center, C3=(−3,−6) and radius r3=3
From the given three circle equations we can observe circles have the same radius r=3

In the above diagram, the radius of the circle touches all the circles represented by:
⇒

CP = CC 1 + C 1 P = CC 1 + 3 = CC 2 + 3 = CC 3 + 3

Let us assume C(h,k) to be the center of the required circle.
Then,

CC 1 = CC 2 = CC 3

or

{(CC 1)}^{2} = {(CC 2)}^{2} = {(CC 3)}^{2}

We get the equation as:

\Rightarrow {(h - 0)}^{2} + {(k - 2)}^{2} = {(h + 6)}^{2} + {(k + 2)}^{2} = {(h + 3)}^{2} + {(k + 6)}^{2}

⇒ Applying, the arithmetic identity,

{(a + b)}^{2} = a^{2} + 2 ab + b^{2}

in the equation (i) as:

\Rightarrow (h^{2}) + (k^{2} - 4 k + 4) = (h^{2} + 12 h + 36) + (k^{2} + 4 k + 4) = (h^{2} + 6 h + 9) + (k^{2} + 12 k + 36)

\Rightarrow (h^{2} + k^{2}) - 4 k + 4 = (h^{2} + k^{2}) + 12 h + 4 k + 40 = (h^{2} + k^{2}) + 6 h + 12 k + 45

\Rightarrow - 4 k + 4 = 12 h + 4 k + 40 = 6 h + 12 k + 45

By equating pair of equations we get:
⇒ 6h+16k+41= 0-- (i)
⇒ And 6h−8k−5= 0--(ii)
From equation (i)
⇒6h+16k+41= 0
⇒6h= −16k−41

\Rightarrow h = \frac{- 16 k - 41}{6} - - - - (iii)

Substitute the value of h from equation (iii) in equation (ii), we get
⇒6h−8k−5=0
⇒8k=6h−5

\Rightarrow 8 k = 6 (\frac{- 16 k - 41}{6}) - 5

⇒8k=−16k−46
⇒24k=−46

\Rightarrow k = - \frac{46}{24} = - \frac{23}{12}

Solving for the value of h we get:

\Rightarrow h = \frac{- 16 k - 41}{6}

\Rightarrow 6 h = - 16 (- \frac{23}{12}) - 41

\Rightarrow 6 h = \frac{4 \times 23}{3} - 41

\Rightarrow 18 h = 92 - 123

\Rightarrow h = - \frac{31}{18}

Hence, by solving equation (i) and (ii) we get

h = - \frac{31}{18} and  k = - \frac{23}{12}

Now we will find the radius of the required circle by substituting the value of h and k in the equation

\Rightarrow CC 1 = \sqrt{{(0 + h)}^{2} + {(2 + k)}^{2}} as : \Rightarrow CC 1 = \sqrt{{(0 + \frac{31}{18})}^{2} + {(2 + \frac{23}{12})}^{2}} = \frac{5}{36} \sqrt{949}

Since,

CP = CC 1 + 3 = \frac{5}{36} \sqrt{949} + 3

The radius of the required circle is

CP = \frac{5}{36} \sqrt{949} + 3

Hence the required equation of the circle is:

\Rightarrow {(x - h)}^{2} + {(y - k)}^{2} = r^{2}

{(x + \frac{31}{18})}^{2} + {(y + \frac{23}{12})}^{2} = {(\frac{5}{39} \sqrt{949} + 3)}^{2}

Detailed Solution