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for what value of non negative integer a will the e system x²-y²=0,(x-a)²+y²=1 have exactly three real solutions
given = x²-y²=0 (x-k)²+y²=1 => x²- 2kx + k²+y² =1 by adding this two equations we will get,2x² - 2kx + k²= 0 and now we all know that if the discriminant is zero i.e., b² + 4ac=0 then the quadratic equation has one solution. Then, 4k² - 8k = 0 4k²=8k 4k = 8 k = 8/4 = 2 so the correct answer is a) 2
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