Find the value of k if x +y-5=0 and 2x+ky-8=0 are conjugate with respect to the circle x square + y square -2x-2y-1=0
Answers
Given : x +y-5=0 and 2x+ky-8=0 are conjugate with respect to the circle x² + y² - 2x - 2y - 1 = 0
To find : Value of k
Solution:
l₁x + m₁y + n₁ = 0
& l₂x + m₂y + n₂ =0
Conjugate to circle x² + y² = r²
if r²(l₁l₂ + m₁m₂) = n₁n₂
x² + y² - 2x - 2y - 1 = 0
=> (x - 1)² - 1 + (y - 1)² - 1 - 1 = 0
=> (x - 1)² + (y - 1)² = (√3)²
=> X² + Y² = (√3)²
X = x - 1 Y = y - 1
=> x = X + 1 & y = Y + 1
putting these in
x +y-5=0 and 2x+ky-8=0
we get
X + 1 + Y + 1 - 5 = 0 & 2(X + 1) + k(Y + 1) - 8 = 0
=> X + Y - 3 = 0 & 2X + kY + k - 6 = 0
X² + Y² = (√3)²
(√3)² (1 * 2 + 1 * k ) = (-3)(k - 6)
=> 3(2 + k) = 3(6 - k)
=> 2 + k = 6 - k
=> 2k = 4
=> k = 2
value of k = 2 if x +y-5=0 and 2x+ky-8=0 are conjugate with respect to the circle x² + y² - 2x - 2y - 1 = 0
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