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10-1.Circle and System of Circles
hard
If the circles ${x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}$ and ${(x - 4)^2} + {(y - 7)^2} = 36$ intersect at two distinct points, then
A
$0 < r < 1$
B
$1 < r < 11$
C
$r>11$
D
$r=11$
(JEE MAIN-2019)
Solution
${x^2} + {y^2} – 16x – 20y + 164 = {r^2}$
$A\left( {8,10} \right),{R_1} = r$
${\left( {x – 4} \right)^2} + {\left( {y – 7} \right)^2} = 36$
$B\left( {4,7,} \right),{R_2} = 6$
$\left| {{R_1} – {R_2}} \right| < AB < {R_1} + {R_2}$
$ \Rightarrow 1 < r < 11$
Standard 11
Mathematics