Condition that circle {(x - 3)^2} + {(y - 4)^2} = {r^2} lies entirely within circle {x^2} + {y^2} =

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10-1.Circle and System of Circles

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The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is

A

$R + r \le 7$

B

${R^2} + {r^2} < 49$

C

${R^2} - {r^2} < 25$

D

$R - r > 5$

Solution

(d) Clearly, ${r_1} – {r_2} > {C_1}{C_2}$

${r_1} = R,$ ${C_1}(3,\,4); $

${r_2} = r;$ ${C_2}(0,\,0)$

$R – r > \sqrt {{{(3 – 0)}^2} + {{(4 – 0)}^2}} $

==> $R -r > 5.$

Standard 11

Mathematics

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