Question

For which value of non-negative 'a' will the system \(x^2 - y^2 = 0\), \((x-a)^2 + y^2 = 1\) have exactly three real solutions?

• A. \(-\sqrt{2}\)
• B. 1
• C. \(\sqrt{2}\)
• D. 2

Hint:

x^2 = y^2 forces y = x or y = -x; substitute into the circle equation to get a quadratic in x, and work out when it must have a root at x = 0 to give an odd (three) total count of solutions.

Answer 1

The Correct Option is B

Step 1: Simplify the first equation.
\(x^2 - y^2 = 0\) factors as \((x-y)(x+y)=0\), so \(y = x\) or \(y = -x\). This is a pair of straight lines through the origin, each at 45 degrees to the axes.

Step 2: Substitute into the circle equation.
Since \(y^2 = x^2\) on both lines, the second equation becomes the same for either choice of sign:
\[ (x-a)^2 + x^2 = 1 \implies 2x^2 - 2ax + (a^2-1) = 0 \]

Step 3: Count solutions produced by each root.
For every root \(x=r\) of this quadratic, we get points on the line \(y=x\) and \(y=-x\): the pair \((r, r)\) and the pair \((r, -r)\). If \(r \ne 0\), these are two distinct solutions to the original system. If \(r = 0\), the two pairs both collapse to the single point \((0,0)\), giving only one solution.

Step 4: Work out when the total is exactly 3.
A quadratic has at most 2 roots. To end up with exactly 3 solutions to the system, we need one root to be \(x=0\) (contributing 1 solution) and the other root to be nonzero and distinct from it (contributing 2 solutions), for a total of \(1+2=3\).

Step 5: Force x = 0 to be a root.
Plug \(x=0\) into \(2x^2-2ax+(a^2-1)=0\):
\[ a^2 - 1 = 0 \implies a = \pm 1 \]
Since a must be non-negative, \(a = 1\).

Step 6: Check the other root is nonzero and distinct.
With \(a=1\): \(2x^2 - 2x = 0 \implies 2x(x-1)=0 \implies x = 0 \text{ or } x = 1\). The two roots are 0 and 1, distinct, and the second one is nonzero. So the system has exactly \((0,0)\) from \(x=0\), plus \((1,1)\) and \((1,-1)\) from \(x=1\): three solutions in total, as required.

Final Answer:
\[ \boxed{a = 1} \]