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
Answers
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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