The equation of the circle of minimum radius which contains the three circles x2 + y2 – 4y – 5 = 0, x2 + y2 + 12x + 4y + 31 = 0 and x2 + y2 + 6x + 12y + 36 = 0 is ...........

C be the centre of the circle passing through the centers of the given circlesC=−3118,−2312radius = CP=CC1+C1P=536949+3C1 - center of 1st circle C1p = radius of 1st circle

The equation of the circle of minimum radius which contains the three circles x2 + y2 – 4y – 5 = 0, x2 + y2 + 12x + 4y + 31 = 0 and x2 + y2 + 6x + 12y + 36 = 0 is ...........

C be the centre of the circle passing through the centers of the given circlesC=−3118,−2312radius = CP=CC1+C1P=536949+3C1 - center of 1st circle C1p = radius of 1st circle

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The equation of the circle of minimum radius which contains the three circles x² + y² – 4y – 5 = 0, x² + y² + 12x + 4y + 31 = 0 and x² + y² + 6x + 12y + 36 = 0 is ...........

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${(x - \frac{31}{18})}^{2} + {(y - \frac{23}{12})}^{2} = {(3 - \frac{5}{36} \sqrt{949})}^{2}$

${(x + \frac{31}{18})}^{2} + {(y + \frac{23}{12})}^{2} = 1$

${(x + \frac{23}{12})}^{2} + {(y + \frac{31}{18})}^{2} = {(3 + \frac{5}{36} \sqrt{949})}^{2}$

answer is D.

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Detailed Solution

C be the centre of the circle passing through the centers of the given circles
$C = (\frac{- 31}{18}, \frac{- 23}{12})$
radius = $CP = {CC}_{1} + C_{1} P = \frac{5}{36} \sqrt{949} + 3$
C₁ - center of 1st circle C₁p = radius of 1st circle