The tangent at any point to the circle `x^(2)+y^(2)=r^(2)` meets the coordinate axes at A and B. If the lines drawn parallel to axes through A and B meet at P then locus of P is

Options

A. `1/(x^(2))+1/(y^(2))=1/(r^(2))`
B. `1/(x^(2))+1/(y^(2))=r^(2)`
C. `1/(x^(2))-1/(y^(2))=r^(2)`
D. `x^(2)+y^(2)=2r^(2)`

Answer

To find the locus of point P, we will follow these steps:

### Step 1: Understand the Circle and Tangent
The equation of the circle is given as:
\[ x^2 + y^2 = r^2 \]
We need to find the tangent line at any point on this circle.

### Step 2: Write the Equation of the Tangent
The equation of the tangent line at a point \((x_1, y_1)\) on the circle can be expressed as:
\[ y = mx + c \]
where \(m\) is the slope of the tangent and \(c\) is the y-intercept.

### Step 3: Condition for Tangency
For the line to be a tangent to the circle, it must satisfy the condition of tangency:
\[ c = \pm r \sqrt{1 + m^2} \]

### Step 4: Find Coordinates of Points A and B
1. **Point A (where the tangent meets the y-axis)**: Set \(x = 0\) in the tangent equation:
\[ A(0, c) \]
2. **Point B (where the tangent meets the x-axis)**: Set \(y = 0\) in the tangent equation:
\[ B\left(-\frac{c}{m}, 0\right) \]

### Step 5: Coordinates of Point P
Point P is defined as the intersection of the lines drawn through A and B parallel to the axes:
- The x-coordinate of P is the x-coordinate of B:
\[ h = -\frac{c}{m} \]
- The y-coordinate of P is the y-coordinate of A:
\[ k = c \]

### Step 6: Substitute for c
From the condition of tangency, we know:
\[ k = c = \pm r \sqrt{1 + m^2} \]

### Step 7: Express h in Terms of k
Substituting \(c\) into the equation for \(h\):
\[ h = -\frac{k}{m} \]

### Step 8: Square Both Sides
1. From \(k = \pm r \sqrt{1 + m^2}\), square both sides:
\[ k^2 = r^2(1 + m^2) \]
2. From \(h = -\frac{k}{m}\), square both sides:
\[ h^2 = \frac{k^2}{m^2} \]

### Step 9: Substitute \(m^2\)
Using \(1 + m^2 = \frac{k^2}{r^2}\), we can express \(m^2\) as:
\[ m^2 = \frac{k^2}{r^2} - 1 \]

### Step 10: Substitute Back
Substituting \(m^2\) into the equation for \(h^2\):
\[ h^2 = \frac{k^2}{\left(\frac{k^2}{r^2} - 1\right)} \]

### Step 11: Simplify the Equation
Cross-multiplying and simplifying gives:
\[ h^2 k^2 = h^2 r^2 + k^2 r^2 \]

### Step 12: Final Equation
Replace \(h\) and \(k\) with \(x\) and \(y\):
\[ x^2 y^2 = x^2 r^2 + y^2 r^2 \]
Dividing through by \(x^2 y^2\):
\[ 1 = \frac{r^2}{y^2} + \frac{r^2}{x^2} \]

### Conclusion
The locus of point P is given by:
\[ \frac{1}{y^2} + \frac{1}{x^2} = \frac{1}{r^2} \]