Question

If (1, 1), (k, 2) are conjugate points with
respect to the circle
x² + y2 +8x +2y +3 = 0
then k=
- 12
- 1217
- 12/5
-4​

Answer 1

Answer:

-1217

Step-by-step explanation:

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Answer 2

Given :   (1, 1), (k, 2) are conjugate points with  respect to the circle

x² + y2 +8x +2y +3 = 0  

To Find :  Value of k

Solution:

x² + y² + 8x + 2y + 3 = 0

=> (x + 4)² - 16 + ( y + 1)² - 1  + 3 = 0

=> (x + 4)² + ( y + 1)² = 14

x + 4 = X  

y + 1 =  Y  

=> (X)² + ( Y)² = 14

x + 4 = X     => 1  + 4 = X => X = 5

y + 1 = Y  =>  1  + 1  = Y => Y = 2

(1, 1) is   (5 , 2)

(k , 2) is ( k + 4 , 3)

x₁ , y₁   and x₂ , y₂ are conjugate to circle x² + y² = r²

if  x₁.x₂ + y₁.y₂  = r₂

=> 5 (k + 4)  + 2(3)  = 14

=> 5k + 20 + 6 = 14

=> 5k  = -12

=> k = - 12/5

or Simply

(x + 4)² + ( y + 1)² = 14

=> (x + 4)(x + 4) + (y + 1)(y + 1) = 14

=> (1 + 4)(k + 4) + (1 + 1)(2 + 1)  = 14

=> 5k + 20 + 6 = 14

=> k = -12/5

Learn More:

Find the value of k if x +y-5=0 and 2x+ky-8=0 are conjugate with ...

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