A) \[c=r({{\lambda }^{2}}+{{\mu }^{2}})\]
B) \[r=c\,({{\lambda }^{2}}+{{\mu }^{2}})\]
C) \[{{c}^{2}}={{r}^{2}}({{\lambda }^{2}}+{{\mu }^{2}})\]
D) \[{{r}^{2}}={{c}^{2}}({{\lambda }^{2}}+{{\mu }^{2}})\]
Correct Answer: C
Solution :
Polar is\[\lambda x+\mu y+c=0\]. The condition of tangency \[p=r\]gives the result (c).You need to login to perform this action.
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