Question

31. If the circle x² + y2 +6x-2y+k=0 bisects
the circumference of the circle
x² + y2 + 2x -6y -15 =0 then k=
(1)21 (2)-21 (3) 23 (4)-23​

Answer 1

Answer:

k = -23  ......  option (4)

Use parametric representation of Circle 2, two end points of its diameter will lie on cicle C1. Substitute in equation and get the answer.

Step-by-step explanation:

Circle C1:  x²+y²+6x-2y + k = , or ,   (x+3)² + (y-1)² = 10 - k  ---(1)

C2:  x²+y²+2x -6y-15=0, or, (x+1)²+(y-3)² = 5²    --- (2)

=> Circle C2:  x = -1 + 5 Cos θ ,   y = 3 + 5 Sin θ   --- (3)   Parametric representation.

C1 intersects C2 at two points A & B (lying on both C1 & C2), where AB is a diameter of C2. The center (-1,3) is the mid point of A & B.

Let A be (-1+5Cosθ, 3+5 Sin θ),    then B = (-1-5cosθ, 3-5 Sinθ)  ... Easy to get it.

Both lie on C1.  So substitute it in (1).

So   (2+5 Cos θ)² + (2+5 Sin θ)² = 10-k   ---(5)

and,  (2-5 Cos θ)² + (2-5 Sin θ)² = 10-k   --- (6)

=>  20 (cosθ+sinθ) = - 20 (cosθ+sinθ)

=>  Cos θ + sin θ = 0  

=> Cosθ = 1/√2 , Sinθ = -1/√2         or, the other way signs reversed.

So points A = (-1+ 5/√2, 3-5/√2)  and B = (-1- 5/√2, 3+5/√2)

A lies on C1.  

So   (2+5/√2)² + (2-5/√2)² = 10-k

=>  33 =  10 - k

=> k = - 23

Answer 2

$\huge\textbf{Answer}$

According to to given question:-

• Consider Circle A1:-

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x²+y²+6x-2y + k = , or , (x+3)² + (y-1)² = 10 - k

• This is equation - (1)

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C2: x²+y²+2x -6y-15=0, or, (x+1)²+(y-3)² = 5²

• This is equation - (2)

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• Circle A2:-

x = -1 + 5 Cos θ and y = 3 + 5 Sin θ

• This is equation - (3)

• This method is called Parametric representation

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• Let this be A1

=(-1+5Cosθ, 3+5 Sin θ)

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• Let this be A2

=(-1-5cosθ, 3-5 Sinθ)

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Substituting values form equation - (1):-

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(2+5 Cos θ)² + (2+5 Sin θ)² = 10-k

• This is equation - (5)

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(2-5 Cos θ)² + (2-5 Sin θ)² = 10-k

• This be equation - (6)

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As per the given steps below:-

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20 (cosθ+sinθ) (=) - 20 (cosθ+sinθ)

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• Cos θ + sin θ = 0
• Cosθ = 1/√2 , Sinθ = -1/√2

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Therefore, The points A equals the below given equations

• (-1+ 5/√2, 3-5/√2)
• B = (-1- 5/√2, 3+5/√2)
• Hence, A lies on C1.

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(2+5/√2)² + (2-5/√2)² = 10-k

33 = 10 - k

k = - 23

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Therefore, the value of [ k = -23]

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