A) \[{{t}_{1}}{{t}_{2}}{{t}_{3}}{{t}_{4}}=-1\]
B) \[{{t}_{1}}{{t}_{2}}{{t}_{3}}{{t}_{4}}=1\]
C) \[{{t}_{1}}{{t}_{3}}={{t}_{2}}{{t}_{4}}\]
D) \[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}={{c}^{2}}\]
Correct Answer: B
Solution :
[b] Let the points lie on the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+2fy+k=0,\] Then \[{{c}^{2}}t_{i}^{2}+\frac{{{c}^{2}}}{t_{i}^{2}}+2ge{{t}_{i}}+2f\frac{c}{{{t}_{i}}}+k=0\] \[\Rightarrow {{c}^{2}}t_{i}^{4}+2gct_{i}^{3}+kt_{i}^{3}+2fc{{t}_{i}}+{{c}^{2}}=0\] Its roots are \[{{t}_{1}},{{t}_{2}},{{t}_{3}},{{t}_{4}}\] so \[{{t}_{1}}{{t}_{2}}{{t}_{3}}{{t}_{4}}=\frac{{{c}^{2}}}{{{c}^{2}}}=1\] Also, \[{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{4}}=-\frac{2gc}{{{c}^{2}}}=-\frac{2g}{c}\]
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