Question

Two circles are given as x^(2)+y^(2)+14x-6y+40=0 and x^(2)+y^(2)-2x+6y+7=0 with their centres as C_(1) and C_(2) .If equation of another circle whose centre C_(3) lies on the line 3x+4y-16=0 and touches the circle C_(1) externally and also C_(1)C_(2)+C_(2)C_(3) is minimum,is x^(2)+y^(2)+ax+by+c=0 then the value of (a+b+c) is​

Answer 1

Given : Two circles are given as x^(2)+y^(2)+14x-6y+40=0 and x^(2)+y^(2)-2x+6y+7=0 with their centres as C_(1) and C_(2)

equation of another circle x^(2)+y^(2)+ax+by+c=0 whose centre C_(3) lies on the line 3x+4y-16=0 and touches the circle C_(1) externally

C_(1)C_(2)+C_(2)C_(3) is minimum,

To Find : the value of (a+b+c)  

Solution:

x²+y²+14x-6y+40=0

=>  (x + 7)² - 49 + (y - 3)² - 9 + 40 = 0

=>  (x + 7)² + (y - 3)² = 18

C₁ = (- 7 , 3)

x²+y²-2x+6y+7=0

=>  (x -1)² - 1 + (y + 3)² - 9 + 7 = 0

=>  (x - 1)² + (y + 3)² = 3

C₂ = (1 , -3)

C₃  lies on the line 3x+4y-16=0

C₁C₂ +  C₂C₃ is minimum

C₁C₂  is fixed hence C₂C₃ is minimum

C₂C₃ is minimum if  C₂C₃ ⊥ (3x + 4y - 16) = 0

Hence  y + 3= (4/3) (x - 1)

=> 3y + 9 = 4x - 4

=> 4x = 3y + 13

   3x+4y-16=0

x = 4  , y = 1    

C₃  = ( 4 , 1)

circle C₃  touches  externally circle C₁

Hence C₁C₃  is a straight line

C₁ = (- 7 , 3)   , C₃  = ( 4 , 1)

Distance =  √125 = 5√5

radius of C₁  =3√2

Hence radius of C₃ = ( 5√5 - 3√2)

Equation of Circle C₃ =  

( x - 4)² + (y - 1)² = ( 5√5 - 3√2)²

=> x²  - 8x + 16  + y² - 2y + 1  =  125 + 18  - 30√10

=> x²   + y² - 8x  - 2y  -126 +  30√10 = 0

a = - 8

b = - 2

c = -126 + 30√10

a + b + c  =  - 136 + 30√10

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