Tangents `PT_1, and PT_2`, are drawn from a point P to the circle `x^2 +y^2=a^2`. If the point P lies on the line `Px +qy + r = 0`, then the locus of the centre of circumcircle of the triangle `PT_1T_2` is

Options

A. `px +qy = r`
B. `(x-p)^(2)+(y-q)^(2) =r^(2)`
C. `px +qy =(r)/(2)`
D. `2px +2qy +r = 0`

Explanation

To solve the problem, we need to find the locus of the center of the circumcircle of triangle \( PT_1T_2 \) formed by the tangents from point \( P \) to the circle defined by the equation \( x^2 + y^2 = a^2 \). Here’s a step-by-step solution:

### Step 1: Understand the Circle and Tangents
The equation of the circle is given as:
\[ x^2 + y^2 = a^2 \]
This represents a circle centered at the origin \( (0, 0) \) with radius \( a \).

### Step 2: Identify the Point P
The point \( P \) lies on the line given by:
\[ px + qy + r = 0 \]
This means that the coordinates of point \( P \) satisfy this linear equation.

### Step 3: Use the Condition for Tangents
From point \( P \), two tangents \( PT_1 \) and \( PT_2 \) can be drawn to the circle. The length of the tangents from point \( P \) to the circle can be expressed using the formula:
\[ PT^2 = OP^2 - r^2 \]
where \( OP \) is the distance from the origin to point \( P \) and \( r \) is the radius of the circle.

### Step 4: Find the Coordinates of Point P
Let \( P(h, k) \) be the coordinates of point \( P \). The distance \( OP \) is given by:
\[ OP = \sqrt{h^2 + k^2} \]
Thus, the condition for the tangents can be expressed as:
\[ h^2 + k^2 - a^2 \geq 0 \]

### Step 5: Locus of the Circumcenter
The circumcenter of triangle \( PT_1T_2 \) lies on the line that is the perpendicular bisector of the segment \( T_1T_2 \). The circumcenter can be expressed in terms of the coordinates of \( P \) and the tangents.

### Step 6: Express the Equation of the Locus
The circumcenter \( O_c \) can be found using the relationship:
\[ x^2 + y^2 - hx - ky = 0 \]
This can be rewritten as:
\[ x^2 + y^2 - hx - ky + a^2 = 0 \]
Substituting \( h \) and \( k \) in terms of \( P \):
\[ px + qy + r = 0 \]
we can express \( h \) and \( k \) in terms of \( x \) and \( y \).

### Step 7: Eliminate h and k
From the line equation, we can express \( h \) and \( k \) as:
\[ h = 2x, \quad k = 2y \]
Substituting these into the equation gives:
\[ p(2x) + q(2y) + r = 0 \]
which simplifies to:
\[ 2px + 2qy + r = 0 \]

### Conclusion
Thus, the locus of the center of the circumcircle of triangle \( PT_1T_2 \) is given by:
\[ 2px + 2qy + r = 0 \]