If the circles x^2+y^2-16x-20y+164=r^2 (r0) | vedclass

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(B) The given circles are $S_1: x^2+y^2-16x-20y+164-r^2=0$ and $S_2: x^2+y^2-8x-14y+29=0$. For $S_1$,the center $C_1 = (8, 10)$ and radius $r_1 = \sqr

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The given circles are $S_1: x^2+y^2-16x-20y+164-r^2=0$ and $S_2: x^2+y^2-8x-14y+29=0$. For $S_1$,the center $C_1 = (8, 10)$ and radius $r_1 = \sqrt{8^2+10^2-(164-r^2)} = \sqrt{64+100-164+r^2} = \sqrt{r^2} = r$ (since $r>0$). For $S_2$,the center $C_2 = (4, 7)$ and radius $r_2 = \sqrt{4^2+7^2-29} = \sqrt{16+49-29} = \sqrt{36} = 6$. The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(8-4)^2+(10-7)^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$. Since the circles intersect at two distinct points,the condition is $|r_1-r_2| Substituting the values,we get $|r-6| From $5  -1$. Since $r>0$,this implies $r > 0$. From $|r-6| Combining these,$r \in (1, 11)$. The maximum integral value of $r$ is $10$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

If the circles $x^2+y^2-16x-20y+164=r^2$ $(r>0)$ and $x^2+y^2-8x-14y+29=0$ intersect in two distinct points,then the maximum possible integral value of $r$ is

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