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Difference of Two Sets

The circle ...

Question

The circle $x^{2} + y^{2} - 6 x - 4 y + 9 = 0$ bisects the circumference of the circle $x^{2} + y^{2} - (λ + 4) x - (λ + 2) y + (5 λ + 3) = 0$ if $λ$ is equal to

0

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Solution

The correct option is B $0$
Since, $C_{1}$ bisects the circumference of $C_{2}$
Then $C_{2}$ passes through the center of $C_{1}$ , i.e $(3, 2)$
$\Rightarrow 3^{2} + 2^{2} - (λ + 4) 3 - (λ + 2) 2 + (5 λ + 3) = 0$
$\Rightarrow λ = 0$

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Similar questions

The circle $x^{2} + y^{2} - 6 x - 4 y + 9 = 0$ bisects the circumference of the circle $x^{2} + y^{2} - (λ + 4) x - (λ + 2) y + (5 λ + 3) = 0 i f λ$ is equal to

Q. If the circle

x^{2} + y^{2} + 4 x - 6 y + c = 0

bisects the circumference of the circle

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

, then c =

Q. For what value of

λ

, the line

4 x + 3 y + λ = 0

will be tangent to the circle

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

Q. The line

x + 2 y + 3 = 0

cuts the circle

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

at points

P

and

Q

and the line

2 x + 3 y + λ = 0

cuts the circle

x^{2} + y^{2} + 6 x + 2 y - 7 = 0

at points

R

and

S .

P,

Q,

R

and

S

are concyclic, then the value of

λ

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The circle x2+y2−6x−4y+9=0 bisects the circumference of the circle x2+y2−(λ+4)x−(λ+2)y+(5λ+3)=0 if λ is equal to

The correct option is B 0Since, C1 bisects the circumference of C2Then C2 passes through the center of C1, i.e (3,2)⇒32+22−(λ+4)3−(λ+2)2+(5λ+3)=0⇒λ=0

The circle $x^{2} + y^{2} - 6 x - 4 y + 9 = 0$ bisects the circumference of the circle $x^{2} + y^{2} - (λ + 4) x - (λ + 2) y + (5 λ + 3) = 0$ if $λ$ is equal to

The correct option is B $0$
Since, $C_{1}$ bisects the circumference of $C_{2}$
Then $C_{2}$ passes through the center of $C_{1}$ , i.e $(3, 2)$
$\Rightarrow 3^{2} + 2^{2} - (λ + 4) 3 - (λ + 2) 2 + (5 λ + 3) = 0$
$\Rightarrow λ = 0$