If the circles {x^2}, + {y^2}, - 16x, - | vedclass

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Question

${x^2} + {y^2} - 16x - 20y + 164 = {r^2}$

$A\left( {8,10} \right),{R_1} = r$

${\left( {x - 4} \right)^2} + {\left( {y - 7} \right)^2} = 36$

$

Vedclass · Accepted Answer

$1 < r < 11$

Vedclass · Answer

${x^2} + {y^2} - 16x - 20y + 164 = {r^2}$

$A\left( {8,10} \right),{R_1} = r$

${\left( {x - 4} \right)^2} + {\left( {y - 7} \right)^2} = 36$

$B\left( {4,7,} \right),{R_2} = 6$

$\left| {{R_1} - {R_2}} \right| < AB < {R_1} + {R_2}$

$ \Rightarrow 1 < r < 11$

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

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