11{\color{red}a_{1} \bar{a_{2}} +a_{2}\bar{a_{1}}=b_{1}+b_{2} } ANSWER[6]
use \left|a_{1} \right|^{2}-b_{1}+\left|a_{2} \right|^{^{2}}-b_{2}=\left|a_{2}-a_{1} \right|^{2}
u will get the answer
What is condition that circles z\bar{z}+\bar{a_{1}}z+a_{1}\bar{z}+b_{1}=0
and z\bar{z}+\bar{a_{2}}z+a_{2}\bar{z}+b_{2}=0
where b_{1},b_{2} \epsilon R
intersect orthogonaly???
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3 Answers
11
24haan haan heek hai..........it was just for practice........and u should that u were good enuf.......[1]