The equation of the line that touches the curves `y=x|x|` and `x^2+(y-2)^2=4` , where `x!=0,` is:

Options

A. `y=4sqrt5x+20`
B. `y=4sqrt3x-12`
C. `y=0`
D. `y=-4sqrt5x-20`

Answer

To find the equation of the line that touches the curves \( y = x|x| \) and \( x^2 + (y - 2)^2 = 4 \) (where \( x \neq 0 \)), we can follow these steps:

### Step 1: Analyze the curves
The first curve \( y = x|x| \) can be rewritten as:
- \( y = x^2 \) for \( x \geq 0 \)
- \( y = -x^2 \) for \( x < 0 \)

The second curve \( x^2 + (y - 2)^2 = 4 \) represents a circle with center at \( (0, 2) \) and radius \( 2 \).

### Step 2: Find the tangent line to \( y = x^2 \)
Assume the equation of the tangent line is given by:
\[ y = mx + c \]
For this line to be tangent to the curve \( y = x^2 \), they must have a common solution. Thus, we set:
\[ x^2 = mx + c \implies x^2 - mx - c = 0 \]
For this quadratic equation to have exactly one solution, the discriminant must be zero:
\[ (-m)^2 - 4(1)(-c) = 0 \implies m^2 + 4c = 0 \implies c = -\frac{m^2}{4} \]

### Step 3: Find the distance from the center of the circle to the tangent line
The distance \( d \) from the center of the circle \( (0, 2) \) to the line \( y = mx + c \) is given by:
\[ d = \frac{|2 - c|}{\sqrt{1 + m^2}} \]
Since this distance must equal the radius of the circle (which is \( 2 \)):
\[ \frac{|2 - c|}{\sqrt{1 + m^2}} = 2 \]
Squaring both sides gives:
\[ (2 - c)^2 = 4(1 + m^2) \]
Substituting \( c = -\frac{m^2}{4} \):
\[ \left(2 + \frac{m^2}{4}\right)^2 = 4(1 + m^2) \]
Expanding and simplifying:
\[ 4 + m^2 + \frac{m^4}{16} = 4 + 4m^2 \]
\[ \frac{m^4}{16} - 3m^2 = 0 \]
Factoring out \( m^2 \):
\[ m^2\left(\frac{m^2}{16} - 3\right) = 0 \]
This gives \( m^2 = 0 \) or \( m^2 = 48 \). Thus, \( m = 0 \) or \( m = \pm 4\sqrt{3} \).

### Step 4: Find corresponding values of \( c \)
For \( m = 0 \):
\[ c = -\frac{0^2}{4} = 0 \implies y = 0 \]
For \( m = 4\sqrt{3} \):
\[ c = -\frac{(4\sqrt{3})^2}{4} = -12 \implies y = 4\sqrt{3}x - 12 \]
For \( m = -4\sqrt{3} \):
\[ c = -\frac{(-4\sqrt{3})^2}{4} = -12 \implies y = -4\sqrt{3}x - 12 \]

### Step 5: Find the tangent line to \( y = -x^2 \)
Following similar steps for the curve \( y = -x^2 \):
1. Set the tangent line \( y = mx + c \).
2. The quadratic equation becomes:
\[ -x^2 - mx - c = 0 \]
The discriminant must also be zero:
\[ (-m)^2 - 4(-1)(-c) = 0 \implies m^2 - 4c = 0 \implies c = \frac{m^2}{4} \]
3. The distance from the center \( (0, 2) \) to the line is:
\[ \frac{|2 - c|}{\sqrt{1 + m^2}} = 2 \]
Substituting \( c = \frac{m^2}{4} \):
\[ \left(2 - \frac{m^2}{4}\right)^2 = 4(1 + m^2) \]
Expanding and simplifying gives:
\[ 4 - m^2 + \frac{m^4}{16} = 4 + 4m^2 \]
\[ \frac{m^4}{16} - 5m^2 = 0 \]
Factoring gives:
\[ m^2\left(\frac{m^2}{16} - 5\right) = 0 \]
Thus, \( m^2 = 0 \) or \( m^2 = 80 \). Hence, \( m = 0 \) or \( m = \pm 4\sqrt{5} \).

### Step 6: Find corresponding values of \( c \)
For \( m = 0 \):
\[ c = \frac{0^2}{4} = 0 \implies y = 0 \]
For \( m = 4\sqrt{5} \):
\[ c = \frac{(4\sqrt{5})^2}{4} = 20 \implies y = 4\sqrt{5}x + 20 \]
For \( m = -4\sqrt{5} \):
\[ c = \frac{(-4\sqrt{5})^2}{4} = 20 \implies y = -4\sqrt{5}x + 20 \]

### Final Result
The equations of the tangents that touch the curves are:
1. \( y = 0 \)
2. \( y = 4\sqrt{3}x - 12 \)
3. \( y = -4\sqrt{3}x - 12 \)
4. \( y = 4\sqrt{5}x + 20 \)
5. \( y = -4\sqrt{5}x + 20 \)