A) \[-r\sqrt{1+{{m}^{2}}}<c\le 0\]
B) \[0\le c<r\sqrt{1+{{m}^{2}}}\]
C) (a) and (b) both
D) \[-c\sqrt{1-{{m}^{2}}}<r\]
Correct Answer: C
Solution :
Substituting equation of line \[y=mx+c\] in circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] \[{{x}^{2}}+{{(mx+c)}^{2}}={{r}^{2}}\Rightarrow {{(1+m)}^{2}}{{x}^{2}}+2mxc+{{c}^{2}}-{{r}^{2}}=0\] If discriminant is greater than zero; two real values of x will be obtained so, \[{{B}^{2}}>4AC\]. \[4{{m}^{2}}{{c}^{2}}-4({{c}^{2}}-{{r}^{2}})(1+{{m}^{2}})>0\] \[{{r}^{2}}(1+{{m}^{2}})>{{c}^{2}}\] \[0\le c<r\sqrt{1+{{m}^{2}}}\] and \[-r\sqrt{(1+{{m}^{2}})}<c\le 0\].
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