`t_(1),t_(2),t_(3)` are lengths of tangents drawn from a point (h,k) to the circles `x^(2)+y^(2)=4,x^(2)+y^(2)-4x=0andx^(2)+y^(2)-4y=0` respectively further, `t_(1)^(4)=t_(2)^(2)" "t_(3)^(2)+16`. Locus of the point (h,k) consist of a straight line `L_(1)` and a circle `C_(1)` passing through origin. A circle `C_(2)` , which is equal to circle `C_(1)` is drawn touching the line `L_(1)` and the circle `C_(1)` externally.
Equation of `C_(1)` is

Options

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

Answer

To solve the problem, we need to find the equation of the circle \( C_1 \) based on the given conditions. Let's break it down step by step.

### Step 1: Identify the circles and their equations
We have three circles:
1. \( C_1: x^2 + y^2 = 4 \) (Circle with center at the origin and radius 2)
2. \( C_2: x^2 + y^2 - 4x = 0 \) (This can be rewritten as \( (x-2)^2 + y^2 = 4 \), center at (2, 0) and radius 2)
3. \( C_3: x^2 + y^2 - 4y = 0 \) (This can be rewritten as \( x^2 + (y-2)^2 = 4 \), center at (0, 2) and radius 2)

### Step 2: Calculate the lengths of the tangents
The lengths of the tangents from a point \( (h, k) \) to a circle \( x^2 + y^2 = r^2 \) is given by the formula:
\[ t = \sqrt{h^2 + k^2 - r^2} \]

For the circles:
- For \( C_1 \): \( r = 2 \)
\[ t_1 = \sqrt{h^2 + k^2 - 4} \]

- For \( C_2 \): \( r = 2 \) and center at (2, 0)
\[ t_2 = \sqrt{h^2 + k^2 - 4h} \]

- For \( C_3 \): \( r = 2 \) and center at (0, 2)
\[ t_3 = \sqrt{h^2 + k^2 - 4k} \]

### Step 3: Set up the equation based on the given condition
We are given the condition:
\[ t_1^4 = t_2^2 \cdot t_3^2 + 16 \]

Substituting the expressions for \( t_1, t_2, \) and \( t_3 \):
\[ (h^2 + k^2 - 4)^2 = (h^2 + k^2 - 4h)(h^2 + k^2 - 4k) + 16 \]

### Step 4: Simplify the equation
Expanding both sides:
1. Left side:
\[ (h^2 + k^2 - 4)^2 = h^4 + k^4 + 16 - 8h^2 - 8k^2 + 2h^2k^2 \]

2. Right side:
Expanding \( (h^2 + k^2 - 4h)(h^2 + k^2 - 4k) \):
\[ = (h^2 + k^2)^2 - 4h(h^2 + k^2) - 4k(h^2 + k^2) + 16hk \]
Simplifying this gives:
\[ = h^4 + k^4 + 2h^2k^2 - 4h^3 - 4k^3 - 4hk(h + k) + 16hk \]

Setting the left side equal to the right side and simplifying will yield a relationship between \( h \) and \( k \).

### Step 5: Find the locus
The locus of the point \( (h, k) \) will consist of a straight line \( L_1 \) and a circle \( C_1 \).

After simplification, we find:
\[ h^2 + k^2 - 2h - 2k = 0 \]
This can be rearranged to:
\[ h^2 + k^2 - 2h - 2k + 1 = 1 \]
This represents a circle centered at \( (1, 1) \) with radius 1.

### Final Answer
The equation of the circle \( C_1 \) is:
\[ x^2 + y^2 - 2x - 2y = 0 \]