Consider the family of circles x^2+y^2-2x-2la | vedclass

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(C) The equation of the family of circles is $x^2+y^2-2x-8-\lambda(2y)=0$. This is of the form $S+\lambda L=0$,where $S=x^2+y^2-2x-8=0$ and $L=2y=0$.

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The equation of the family of circles is $x^2+y^2-2x-8-\lambda(2y)=0$. This is of the form $S+\lambda L=0$,where $S=x^2+y^2-2x-8=0$ and $L=2y=0$. The fixed points $A$ and $B$ are the intersection points of the circle $S=0$ and the line $L=0$. Substituting $y=0$ into $x^2+y^2-2x-8=0$,we get $x^2-2x-8=0$. Factoring the quadratic equation: $(x-4)(x+2)=0$,which gives $x=4$ and $x=-2$. Thus,the points are $A(4, 0)$ and $B(-2, 0)$. The distance between $A$ and $B$ is $|4 - (-2)| = |6| = 6$. Alternatively,the distance between the points of intersection of $x^2+y^2+2gx+2fy+c=0$ and $y=0$ is $\sqrt{4g^2-4c} = \sqrt{4(-1)^2-4(-8)} = \sqrt{4+32} = \sqrt{36} = 6$. Hence,option $C$ is correct.

Consider the family of circles $x^2+y^2-2x-2\lambda y-8=0$. This family passes through two fixed points $A$ and $B$. Find the distance between these two points.

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