Three concentric circles,of which the biggest | vedclass

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(C) Let the radii of the three concentric circles be $r_1, r_2, r_3$ in $A.P.$ such that $r_3 = 1$ (the largest). Let the common difference be $d > 0$

Vedclass · Accepted Answer

$\left( 0, \frac{2 - \sqrt{2}}{4} \right)$

Vedclass · Answer

(C) Let the radii of the three concentric circles be $r_1, r_2, r_3$ in $A.P.$ such that $r_3 = 1$ (the largest). Let the common difference be $d > 0$. Then $r_3 = 1$,$r_2 = 1 - d$,and $r_1 = 1 - 2d$.For the line $x - y + 1 = 0$ to cut all circles at two real and distinct points,the perpendicular distance from the center $(0, 0)$ to the line must be less than the radius of the smallest circle $r_1$.The perpendicular distance $p = \frac{|0 - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$.Thus,we require $r_1 > p$,which means $1 - 2d > \frac{1}{\sqrt{2}}$.Solving for $d$: $2d $d Since $d > 0$,the interval for $d$ is $\left( 0, \frac{2 - \sqrt{2}}{4} \right)$.

Three concentric circles,of which the biggest is $x^2 + y^2 = 1$,have their radii in $A.P.$ If the line $y = x + 1$ cuts all the circles in real and distinct points,the interval in which the common difference $d$ of the $A.P.$ will lie is:

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