1. For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?$$x^2 - y^2 = 0$$$$(x-k)^2 + y^2 = 1$$
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By: anil on 05 May 2019 02.29 am
From 1st equation we know that $$(x)^2 = y^2 $$ Substituting this in 2nd equation. we get , $$2*x^2 - 2*x*k + k^2-1 =0 $$ and for unique solution $$b^2-4ac=0$$ must satisfy. This is possible only when k = $$sqrt{2}$$
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