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 to the given ellipse \(\frac{(x-1)^2}{16} + \frac{(y-1)^2}{9} = 1\), we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be compared with the standard form of the ellipse equation:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
From the given equation, we identify:
- Center \((h, k) = (1, 1)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 2: Use the property of perpendicular tangents
The locus of the point of intersection of the perpendicular tangents of an ellipse is a circle. The radius of this circle is given by:
\[ \sqrt{a^2 + b^2} \]
Calculating this, we have:
\[ a^2 + b^2 = 16 + 9 = 25 \]
Thus, the radius is:
\[ \sqrt{25} = 5 \]

### Step 3: Write the equation of the locus
The center of the circle is the same as the center of the ellipse, which is \((1, 1)\). Therefore, the equation of the circle is:
\[ (x - 1)^2 + (y - 1)^2 = 5^2 \]
This simplifies to:
\[ (x - 1)^2 + (y - 1)^2 = 25 \]

### Step 4: Expand the equation
Expanding the equation gives:
\[ (x^2 - 2x + 1) + (y^2 - 2y + 1) = 25 \]
Combining like terms, we have:
\[ x^2 + y^2 - 2x - 2y + 2 = 25 \]
Subtracting 25 from both sides results in:
\[ x^2 + y^2 - 2x - 2y - 23 = 0 \]

### Final Answer
The locus of the point of intersection of the perpendicular tangents of the ellipse is given by:
\[ x^2 + y^2 - 2x - 2y - 23 = 0 \]