For the two circles x^2 + y^2 = 16 and x^2 + | vedclass

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Question

Let, ${x^2} + {y^2} = 16$ or${x^2} + {y^2} = {4^2}$

radius of circle ${r_1} = 4\,$ center ${C_1}\left( {0,0} \right)$

we have&nbs

Vedclass · Accepted Answer

no common tangent

Vedclass · Answer

Let, ${x^2} + {y^2} = 16$ or${x^2} + {y^2} = {4^2}$

radius of circle ${r_1} = 4\,$ center ${C_1}\left( {0,0} \right)$

we have ${x^2} + {y^2} - 2y = 0$

 or ${x^2} + {\left( {y - 1} \right)^2} = {1^2}$

Rsdius $1$, center ${C_2}\left( {0,1} \right)$

$\left| {{C_1}{C_2}} \right| = 1$

$\left| {{r_2} - {r_1}} \right| = \left| {4 - 1} \right| = 3$

$\left| {{C_1}{C_2}} \right| > \left| {{r_2} - {r_1}} \right|$

no common tangents for these two circles.

For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 -2y = 0,$ there is/are

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