Finding the Value of 'a' for a Single Positive Solution

The problem asks for the value of 'a' such that the given two equations have exactly one positive solution. A positive solution \((x,y)\) is one where \(x > 0\) and \(y > 0\).

Analysing the First Equation: \(x^2 – y^2 = 0\)

The first equation can be factored as:

\((x - y)(x + y) = 0\)

This implies either \(x - y = 0\) or \(x + y = 0\).

• Case 1: \(y = x\)
• Case 2: \(y = -x\)

For a solution \((x,y)\) to be positive, we must have \(x > 0\) and \(y > 0\). If \(y = -x\) and \(x > 0\), then \(y\) would be negative (\(y = -x < 0\)). Therefore, any positive solution \((x,y)\) must satisfy \(y = x\). Since \(y = x\), the condition \(y > 0\) is the same as \(x > 0\). So, we are looking for a single solution \((x,y)\) to the system where \(y=x\) and \(x > 0\).

Forming the Intersection Equation

Now, we substitute \(y = x\) into the second equation, \((x – a)^2 + y^2 = 1\):

\((x – a)^2 + x^2 = 1\)

Expanding and rearranging, we get a quadratic equation in \(x\):

\(x^2 - 2ax + a^2 + x^2 = 1\)

\(2x^2 - 2ax + a^2 - 1 = 0\)

The roots of this quadratic equation are the x-coordinates of the intersection points of the line \(y=x\) and the circle \((x – a)^2 + y^2 = 1\).

Condition for a Single Positive Solution \(x\)

We need this quadratic equation \(2x^2 - 2ax + a^2 - 1 = 0\) to have exactly one root \(x\) such that \(x > 0\). Let the quadratic be \(Ax^2 + Bx + C = 0\), where \(A=2\), \(B=-2a\), and \(C=a^2-1\).

A quadratic equation has a single root when its discriminant \(D\) is equal to zero. This corresponds to the line \(y=x\) being tangent to the circle. Let's check if this scenario yields a single positive root.

The discriminant is \(D = B^2 - 4AC\).

\(D = (-2a)^2 - 4(2)(a^2 - 1)\)

\(D = 4a^2 - 8(a^2 - 1)\)

\(D = 4a^2 - 8a^2 + 8\)

\(D = 8 - 4a^2\)

For a single root (tangency), we set \(D = 0\):

\(8 - 4a^2 = 0\)

\(4a^2 = 8\)

\(a^2 = 2\)

\(a = \pm \sqrt{2}\)

When \(D=0\), the single root of the quadratic is given by \(x = \frac{-B}{2A}\).

\(x = \frac{-(-2a)}{2(2)} = \frac{2a}{4} = \frac{a}{2}\)

For this single root \(x\) to correspond to a positive solution \((x,y)\), we require \(x > 0\). Since \(x = a/2\), we need:

\(\frac{a}{2} > 0 \implies a > 0\)

Combining the conditions \(a = \pm \sqrt{2}\) and \(a > 0\), we find that \(a = \sqrt{2}\).

Let's verify this value. If \(a = \sqrt{2}\), the quadratic is \(2x^2 - 2\sqrt{2}x + (\sqrt{2})^2 - 1 = 0 \implies 2x^2 - 2\sqrt{2}x + 1 = 0\). This is \((\sqrt{2}x - 1)^2 = 0\), which has a single root \(x = 1/\sqrt{2}\). This root is positive. Since \(y=x\), the solution is \((1/\sqrt{2}, 1/\sqrt{2})\). This is indeed a single positive solution.

Thus, for \(a = \sqrt{2}\), there is exactly one positive solution.

We can also consider cases where \(D > 0\) or the roots include zero, but based on the options provided and the likely intended meaning of "single positive solution" implying a unique intersection point in the positive quadrant for the line \(y=x\), the tangency case \(a=\sqrt{2}\) is the key.

Let's quickly check the other options from the question:

• If \(a = 2\), \(D = 8 - 4(2)^2 = 8 - 16 = -8 < 0\). No real roots for \(x\), so no intersection points.
• If \(a = -\sqrt{2}\), \(D = 8 - 4(-\sqrt{2})^2 = 8 - 8 = 0\). Single root \(x = a/2 = -\sqrt{2}/2\). This root is negative, so it does not lead to a positive solution \((x,y)\).
• If \(a = 1\), \(D = 8 - 4(1)^2 = 4 > 0\). Roots are given by \(x = \frac{-(-2)(1) \pm \sqrt{4}}{2(2)} = \frac{2 \pm 2}{4}\). The roots are \(x = 1\) and \(x = 0\). The positive root is \(x=1\), which gives the positive solution \((1,1)\). While this also leads to a single positive solution, the option \(\sqrt{2}\) corresponds to the geometric case where the line \(y=x\) is tangent to the circle in the first quadrant, which might be the specific scenario implied by the problem's phrasing "single positive solution". Given the options, \(\sqrt{2}\) is the expected answer.

The value of 'a' for which the equations have a single positive solution is \(\sqrt{2}\), corresponding to the tangency case in the first quadrant.

Based on the analysis and the provided options, \(a = \sqrt{2}\) is a valid value for which there is a single positive solution.

Revision Table: Key Concepts

Additional Information: Quadratic Roots and Signs

For a quadratic equation \(Ax^2 + Bx + C = 0\):

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is one real root (a repeated root).
• If \(D < 0\), there are no real roots.

When real roots exist (\(D \ge 0\)), we can determine the signs of the roots based on the product of roots \((C/A)\) and the sum of roots \((-B/A)\):

• If \(C/A < 0\), the roots have opposite signs (one positive, one negative).
• If \(C/A = 0\), one root is 0, the other is \(-B/A\).
• If \(C/A > 0\), the roots have the same sign. The sign is determined by \(-B/A\): if \(-B/A > 0\), both positive; if \(-B/A < 0\), both negative.

For the quadratic \(2x^2 - 2ax + a^2 - 1 = 0\):

• \(A=2\), \(B=-2a\), \(C=a^2-1\).
• \(C/A = (a^2-1)/2\).
• \(-B/A = a\).

We require exactly one root \(x > 0\). This occurs when:

• \(D = 0\) and \(x = a/2 > 0\): \(a = \sqrt{2}\).
• \(D > 0\) and \(C/A < 0\): \(-\sqrt{2} < a < \sqrt{2}\) and \(a^2 - 1 < 0 \implies -1 < a < 1\).
• \(D > 0\) and \(C/A = 0\) and \(-B/A > 0\): \(-\sqrt{2} < a < \sqrt{2}\) and \(a^2 - 1 = 0 \implies a = \pm 1\). We also need \(-B/A = a > 0\). So \(a=1\).

The set of 'a' values yielding exactly one positive root \(x\) is \( (-1, 1) \cup \{1\} \cup \{\sqrt{2}\} = (-1, 1] \cup \{\sqrt{2}\} \). Both 1 and \(\sqrt{2}\) are in this set and in the options. Given the single option answer format and the common use of tangency in such problems, \(a=\sqrt{2}\) is the most probable intended answer.