Consider the family of circles $x^{2} + y^{2} - 2 x - 2 λ y - 8 = 0$ passing through two fixed points $A$ and $B$ . Then the distance between the points $A$ and $B$ is
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Solution

$x^{2} + y^{2} - 2 x - 2 λ y - 8 = 0$
$\Rightarrow (x^{2} + y^{2} - 2 x - 8) - 2 λ y = 0$ , which is of the form of $S + λ L = 0$
All the circles passing through the point of intersection of circle $x^{2} + y^{2} - 2 x - 8 = 0$ and $y = 0$
Solving we get $x^{2} - 2 x + 8 = 0$ or $(x - 4) (x + 2) = 0$
$\Rightarrow A \equiv (4, 0)$ and $B \equiv (- 2, 0)$
$\Rightarrow$ Distance between $A$ and $B$ is $6$ .

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Consider the family of circles x2+y2−2x−2λy−8=0 passing through two fixed points A and B. Then the distance between the points A and B is units

x2+y2−2x−2λy−8=0 ⇒(x2+y2−2x−8)−2λy=0, which is of the form of S+λL=0 All the circles passing through the point of intersection of circle x2+y2−2x−8=0 and y=0 Solving we get x2−2x+8=0 or (x−4)(x+2)=0 ⇒A≡(4,0) and B≡(−2,0) ⇒ Distance between A and B is 6.

Consider the family of circles $x^{2} + y^{2} - 2 x - 2 λ y - 8 = 0$ passing through two fixed points $A$ and $B$ . Then the distance between the points $A$ and $B$ is
units