The locus of point intersection of perpendicular tangents of ellipse `((x-1)^(2))/(16)+((y-1)^(2))/(9)=1` is

Options

A. `x^(2)+y^(2)=25`
B. `x^(2)+y^(2)+2x+2y-23=0`
C. `x^(2)+y^(2)-2x-2y-23=0`
D. none of these

Answer

To find the locus of the point of intersection of perpendicular tangents of the given ellipse \(\frac{(x-1)^2}{16} + \frac{(y-1)^2}{9} = 1\), we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The standard form of the ellipse is given by:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
where \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

For our ellipse:
- Center \((h, k) = (1, 1)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 2: Use the formula for the locus of the intersection of perpendicular tangents
The locus of the point of intersection of perpendicular tangents to an ellipse is given by the equation:
\[ x^2 + y^2 = a^2 + b^2 \]

### Step 3: Calculate \(a^2 + b^2\)
Substituting the values of \(a^2\) and \(b^2\):
\[ a^2 + b^2 = 16 + 9 = 25 \]

### Step 4: Write the equation of the locus
Substituting \(a^2 + b^2\) into the locus equation:
\[ x^2 + y^2 = 25 \]

### Step 5: Final equation
This equation represents a circle centered at the origin with a radius of 5. However, since the ellipse is centered at \((1, 1)\), we need to adjust the equation to reflect the center of the ellipse.

The locus of the point of intersection of the perpendicular tangents is:
\[ (x-1)^2 + (y-1)^2 = 25 \]

### Conclusion
Thus, the locus of the point of intersection of perpendicular tangents of the ellipse is:
\[ (x-1)^2 + (y-1)^2 = 25 \]

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