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 ...........

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 ...........

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

none

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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