The equation of tangent to the circle `x^(2) +y^(2) =a^(2)` which makes with axes a triangle of area `a^(2)` is `:`

Options

A. `y =x +- a sqrt(2)`
B. `y = - x+-a `
C. `y =x +- a `
D. `y = x+- (a)/( sqrt(2))`

Answer

To find the equation of the tangent to the circle \(x^2 + y^2 = a^2\) that forms a triangle with the axes of area \(a^2\), we can follow these steps:

### Step 1: Understand the Circle and Tangent
The given circle has a radius \(a\) and is centered at the origin (0, 0). The equation of a tangent line to the circle can be expressed in the intercept form.

### Step 2: Set Up the Tangent Equation
Let the tangent line cut the x-axis at point \(C(c, 0)\) and the y-axis at point \(D(0, d)\). The equation of the tangent line can be written as:
\[ \frac{y}{d} + \frac{x}{c} = 1 \]
This can be rearranged to:
\[ cy + dx = cd \]

### Step 3: Calculate the Area of Triangle
The area \(A\) of triangle \(OCD\) (formed by the axes and the tangent) is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times c \times d \]
According to the problem, this area is equal to \(a^2\):
\[ \frac{1}{2}cd = a^2 \implies cd = 2a^2 \]

### Step 4: Find the Distance from the Origin to the Tangent
The distance from the origin (0, 0) to the line \(cy + dx - cd = 0\) must equal the radius \(a\). The formula for the distance \(D\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line, we have \(A = d\), \(B = c\), and \(C = -cd\):
\[ D = \frac{|0 + 0 - cd|}{\sqrt{c^2 + d^2}} = \frac{cd}{\sqrt{c^2 + d^2}} \]
Setting this equal to the radius \(a\):
\[ \frac{cd}{\sqrt{c^2 + d^2}} = a \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ \frac{c^2d^2}{c^2 + d^2} = a^2 \]
Cross-multiplying yields:
\[ c^2d^2 = a^2(c^2 + d^2) \]

### Step 6: Substitute \(cd = 2a^2\)
From \(cd = 2a^2\), we can substitute \(d = \frac{2a^2}{c}\) into the equation:
\[ c^2 \left(\frac{2a^2}{c}\right)^2 = a^2\left(c^2 + \left(\frac{2a^2}{c}\right)^2\right) \]
This simplifies to:
\[ \frac{4a^4}{c^2} = a^2\left(c^2 + \frac{4a^4}{c^2}\right) \]

### Step 7: Solve for \(c\) and \(d\)
After simplification, we find that:
\[ c^2 + d^2 = 2cd \]
This implies:
\[ (c - d)^2 = 0 \implies c = d \]
Substituting \(d = c\) into \(cd = 2a^2\) gives:
\[ c^2 = 2a^2 \implies c = d = \sqrt{2}a \]

### Step 8: Write the Tangent Equation
Substituting \(c\) and \(d\) back into the tangent equation:
\[ \frac{y}{\sqrt{2}a} + \frac{x}{\sqrt{2}a} = 1 \]
This simplifies to:
\[ x + y = \sqrt{2}a \]

### Conclusion
Thus, the equation of the tangent to the circle \(x^2 + y^2 = a^2\) that forms a triangle of area \(a^2\) with the axes is:
\[ \boxed{x + y = \sqrt{2}a} \]