Question

Prove that the circle x^2 + y^2 - 6x - 4y + 9 = 0 bisects the circumference

Answer 1

Answer:

Step-by-step explanation:

7xy

Answer 2

$\huge\textbf{Answer}$

According to to given question:-

Consider Circle A1:-

x²+y²+6x-2y + k = 0

or

(x+3)² + (y-1)² = 10 - k

This is equation ______________ (1)

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

This is equation ___________________(2)

Circle A2:-

x = -1 + 5 Cos θ and y = 3 + 5 Sin θ

This is equation ______________(3)

This method is called Parametric representation.

Let this be A1

=(-1+5Cosθ, 3+5 Sin θ)

Let this be A2

=(-1-5cosθ, 3-5 Sinθ)

Substituting values form equation (1):-

(2+5 Cos θ)² + (2+5 Sin θ)² = 10-k

This is equation _____________(5)

(2-5 Cos θ)² + (2-5 Sin θ)² = 10-k

This be equation ____________(6)

As per the given steps below:-

20 (cosθ+sinθ) (=) - 20 (cosθ+sinθ)

Cos θ + sin θ = 0

Cosθ = 1/√2 , Sinθ = -1/√2

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.

(2+5/√2)² + (2-5/√2)² = 10-k

33 = 10 - k

k = - 23

Therefore,

The value of [ k = -23]