A) \[a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}\]
B) \[\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}\]
C) \[\frac{1}{2}(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})\]
D) \[\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\]
Correct Answer: D
Solution :
Let \[(h,\,\,k)\] be the point on the locus, then by the given conditions \[{{(h-{{a}_{1}})}^{2}}+{{(k-{{b}_{1}})}^{2}}={{(h-{{a}_{2}})}^{2}}+{{(k-{{b}_{2}})}^{2}}\] \[\Rightarrow \,\,2h\,({{a}_{1}}-{{a}_{2}})+2k\,({{b}_{1}}-{{b}_{2}})+a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}=0\] \[\Rightarrow \,\,h\,({{a}_{1}}-{{a}_{2}})+k\,({{b}_{1}}-{{b}_{2}})+\frac{1}{2}\,(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})=0\]....(i) Also, since (h, k) lies on the given locus, therefore \[({{a}_{1}}-{{a}_{2}})\,h+({{b}_{1}}-{{b}_{2}})\,k+c=0\] .....(ii) Comparing (i) and (ii), we get \[c=\frac{1}{2}\,(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\].
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