The range of values of a for which the point (a, 4) is outside the circles `x^(2)+y^(2)+10x=0` and `x^(2)+y^(2)-12x+20=0`, is

Options

A. `(-oo, -8) uu(-2,6) uu (6, oo)`
B. (-8, -2)
C. `(-oo, -2)uu(-2, oo)`
D. none of these

Answer

To find the range of values of \( a \) for which the point \( (a, 4) \) is outside the circles given by the equations \( x^2 + y^2 + 10x = 0 \) and \( x^2 + y^2 - 12x + 20 = 0 \), we will follow these steps:

### Step 1: Rewrite the equations of the circles
1. **First Circle**:
The equation \( x^2 + y^2 + 10x = 0 \) can be rewritten by completing the square:
\[ x^2 + 10x + y^2 = 0 \implies (x + 5)^2 + y^2 = 25 \]
This represents a circle centered at \( (-5, 0) \) with a radius of \( 5 \).

2. **Second Circle**:
The equation \( x^2 + y^2 - 12x + 20 = 0 \) can also be rewritten:
\[ x^2 - 12x + y^2 + 20 = 0 \implies (x - 6)^2 + y^2 = 16 \]
This represents a circle centered at \( (6, 0) \) with a radius of \( 4 \).

### Step 2: Determine the conditions for the point \( (a, 4) \) to be outside the circles
For the point \( (a, 4) \) to be outside a circle, the distance from the point to the center of the circle must be greater than the radius of the circle.

1. **For the First Circle**:
The distance from \( (a, 4) \) to the center \( (-5, 0) \) is:
\[ d_1 = \sqrt{(a + 5)^2 + (4 - 0)^2} = \sqrt{(a + 5)^2 + 16} \]
We need this distance to be greater than the radius \( 5 \):
\[ \sqrt{(a + 5)^2 + 16} > 5 \]
Squaring both sides:
\[ (a + 5)^2 + 16 > 25 \implies (a + 5)^2 > 9 \]
Taking square roots:
\[ |a + 5| > 3 \]
This leads to two inequalities:
\[ a + 5 > 3 \quad \text{or} \quad a + 5 < -3 \]
Simplifying gives:
\[ a > -2 \quad \text{or} \quad a < -8 \]

2. **For the Second Circle**:
The distance from \( (a, 4) \) to the center \( (6, 0) \) is:
\[ d_2 = \sqrt{(a - 6)^2 + (4 - 0)^2} = \sqrt{(a - 6)^2 + 16} \]
We need this distance to be greater than the radius \( 4 \):
\[ \sqrt{(a - 6)^2 + 16} > 4 \]
Squaring both sides:
\[ (a - 6)^2 + 16 > 16 \implies (a - 6)^2 > 0 \]
This implies:
\[ a - 6 \neq 0 \implies a \neq 6 \]

### Step 3: Combine the conditions
From the first circle, we have:
- \( a > -2 \) or \( a < -8 \)

From the second circle, we have:
- \( a \neq 6 \)

Combining these conditions, we find that the point \( (a, 4) \) is outside both circles when:
- \( a < -8 \) or \( a > -2 \) (excluding \( a = 6 \))

### Final Answer
Thus, the range of values of \( a \) for which the point \( (a, 4) \) is outside both circles is:
\[ (-\infty, -8) \cup (-2, 6) \cup (6, \infty) \]