Tangents from point P are drawn one of each of th circle `x^(2)+y^(2)-4x-8y+11=0` and `x^(2)+y^(2)-4x-8y+15=0` if the tangents are perpendicular then the locus o P is

Options

A. `x^(2)+y^(2)-4x-8y+6=0`
B. `x+y+10=0`
C. `y^(2)-2(x-8)`
D. `x^(2)+y^(2)-4x-8y-6=0`

Answer

To solve the problem, we need to find the locus of point P from which tangents are drawn to two given circles, and the tangents are perpendicular to each other. Let's break down the solution step by step.

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( x^2 + y^2 - 4x - 8y + 11 = 0 \)
2. \( x^2 + y^2 - 4x - 8y + 15 = 0 \)

### Step 2: Rewrite the equations in standard form
We can rewrite these equations in standard form by completing the square.

For the first circle:
\[ x^2 - 4x + y^2 - 8y + 11 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x-2)^2 - 4 + (y-4)^2 - 16 + 11 = 0 \]
\[ (x-2)^2 + (y-4)^2 - 9 = 0 \]
This gives us the first circle with center \(C_1(2, 4)\) and radius \(r_1 = 3\).

For the second circle:
\[ x^2 - 4x + y^2 - 8y + 15 = 0 \]
Completing the square:
\[ (x-2)^2 - 4 + (y-4)^2 - 16 + 15 = 0 \]
\[ (x-2)^2 + (y-4)^2 - 5 = 0 \]
This gives us the second circle with the same center \(C_2(2, 4)\) and radius \(r_2 = \sqrt{5}\).

### Step 3: Determine the condition for perpendicular tangents
The tangents from point P to the circles are perpendicular if the following condition holds:
\[ OP^2 = r_1^2 + r_2^2 \]
where \(OP\) is the distance from point P to the center of the circles.

### Step 4: Calculate \(r_1^2\) and \(r_2^2\)
We have:
- \(r_1 = 3 \Rightarrow r_1^2 = 9\)
- \(r_2 = \sqrt{5} \Rightarrow r_2^2 = 5\)

Thus,
\[ r_1^2 + r_2^2 = 9 + 5 = 14 \]

### Step 5: Set up the equation for the locus of P
Let \(P(x, y)\) be any point from which tangents are drawn. The distance \(OP\) from point P to the center (2, 4) is given by:
\[ OP^2 = (x - 2)^2 + (y - 4)^2 \]
Setting this equal to \(r_1^2 + r_2^2\):
\[ (x - 2)^2 + (y - 4)^2 = 14 \]

### Step 6: Write the final equation of the locus
The locus of point P is a circle with center (2, 4) and radius \(\sqrt{14}\):
\[ (x - 2)^2 + (y - 4)^2 = 14 \]

### Conclusion
The locus of point P from which the tangents to the two circles are perpendicular is given by the equation:
\[ (x - 2)^2 + (y - 4)^2 = 14 \]