`x + ay =a ^(2) +1` is a tangent to the circle `x ^(2) + y ^(2) =10` for

Options

A. any values of a
B. only one value of a
C. two value of a
D. no value of a.

Answer

To determine the value of \( a \) for which the line \( x + ay = a^2 + 1 \) is a tangent to the circle \( x^2 + y^2 = 10 \), we can follow these steps:

### Step 1: Identify the Circle's Radius
The equation of the circle is given by:
\[ x^2 + y^2 = 10 \]
This can be compared to the standard form \( x^2 + y^2 = r^2 \), where \( r \) is the radius. Thus, we have:
\[ r^2 = 10 \implies r = \sqrt{10} \]

### Step 2: Rewrite the Line Equation
The line equation is given as:
\[ x + ay = a^2 + 1 \]
We can rearrange this into the slope-intercept form \( y = mx + c \):
\[ ay = -x + (a^2 + 1) \implies y = -\frac{1}{a}x + \frac{a^2 + 1}{a} \]
Here, the slope \( m \) of the line is \( -\frac{1}{a} \).

### Step 3: Use the Tangent Condition
For the line to be a tangent to the circle, the distance from the center of the circle (which is at the origin \( (0, 0) \)) to the line must equal the radius \( r \). The distance \( D \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( x + ay - (a^2 + 1) = 0 \), we have:
- \( A = 1 \)
- \( B = a \)
- \( C = -(a^2 + 1) \)

Substituting \( (x_0, y_0) = (0, 0) \):
\[ D = \frac{|1 \cdot 0 + a \cdot 0 - (a^2 + 1)|}{\sqrt{1^2 + a^2}} = \frac{|-(a^2 + 1)|}{\sqrt{1 + a^2}} = \frac{a^2 + 1}{\sqrt{1 + a^2}} \]

### Step 4: Set Distance Equal to Radius
Setting the distance \( D \) equal to the radius \( r \):
\[ \frac{a^2 + 1}{\sqrt{1 + a^2}} = \sqrt{10} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ \frac{(a^2 + 1)^2}{1 + a^2} = 10 \]
Multiplying both sides by \( 1 + a^2 \):
\[ (a^2 + 1)^2 = 10(1 + a^2) \]

### Step 6: Expand and Rearrange
Expanding the left side:
\[ a^4 + 2a^2 + 1 = 10 + 10a^2 \]
Rearranging gives:
\[ a^4 + 2a^2 - 10a^2 + 1 - 10 = 0 \implies a^4 - 8a^2 - 9 = 0 \]

### Step 7: Substitute \( u = a^2 \)
Let \( u = a^2 \):
\[ u^2 - 8u - 9 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2} \]
Calculating the two possible values:
1. \( u = \frac{18}{2} = 9 \)
2. \( u = \frac{-2}{2} = -1 \) (not valid since \( u = a^2 \geq 0 \))

Thus, \( u = 9 \) implies:
\[ a^2 = 9 \implies a = 3 \text{ or } a = -3 \]

### Final Answer
The values of \( a \) for which the line is a tangent to the circle are:
\[ a = 3 \text{ or } a = -3 \]