A and B are two points (0,0) and (3a,0) respectively. Points P and Q are taken on AB such that AP=PQ=QB. Circles are drawn on AP, PQ and QB as diameters. If T be the point from where the sum of the squares of the lengths of tangents to these three circles be `b^(2)` then locus of the point T is

Options

A. `x^(2)+y^(2)-5ax+6a^(2)-b^(2)=0`
B. `x^(2)+y^(2)-ax-b^(2)=0`
C. `x^(2)+y^(2)-3ax+2a^(2)-b^(2)=0`
D. `3(x^(2)+y^(2))-9ax+8a^(2)-b^(2)=0`

Answer

To solve the problem step by step, we will follow the outlined reasoning and calculations from the video transcript.

### Step 1: Identify Points A and B
Given points:
- A = (0, 0)
- B = (3a, 0)

### Step 2: Determine Points P and Q
Since AP = PQ = QB, let:
- AP = PQ = QB = x
- Therefore, the total length AB = AP + PQ + QB = 3x
- Since AB = 3a, we have 3x = 3a, which gives x = a.

Now we can find the coordinates of P and Q:
- P = (AP, 0) = (a, 0)
- Q = (AP + PQ, 0) = (a + a, 0) = (2a, 0)

### Step 3: Define the Circles
We will define the circles with diameters AP, PQ, and QB.

1. **Circle with diameter AP**:
- Center = ((0 + a)/2, (0 + 0)/2) = (a/2, 0)
- Radius = AP/2 = a/2
- Equation: \((x - \frac{a}{2})^2 + y^2 = \left(\frac{a}{2}\right)^2\)
- Simplifying gives: \(x^2 + y^2 - ax = 0\) (denote this as S1)

2. **Circle with diameter PQ**:
- Center = ((a + 2a)/2, (0 + 0)/2) = (3a/2, 0)
- Radius = PQ/2 = a/2
- Equation: \((x - \frac{3a}{2})^2 + y^2 = \left(\frac{a}{2}\right)^2\)
- Simplifying gives: \(x^2 + y^2 - 3ax + \frac{3a^2}{4} = 0\) (denote this as S2)

3. **Circle with diameter QB**:
- Center = ((2a + 3a)/2, (0 + 0)/2) = (5a/2, 0)
- Radius = QB/2 = a/2
- Equation: \((x - \frac{5a}{2})^2 + y^2 = \left(\frac{a}{2}\right)^2\)
- Simplifying gives: \(x^2 + y^2 - 5ax + \frac{25a^2}{4} = 0\) (denote this as S3)

### Step 4: Length of Tangents from Point T
Let T be the point (h, k). The length of the tangents from point T to each circle is given by the formula:
\[ \text{Length} = \sqrt{S} \]
where S is the equation of the circle evaluated at (h, k).

1. For S1:
\[ S1 = h^2 + k^2 - ah \]
\[ \text{Length}^2 = h^2 + k^2 - ah \]

2. For S2:
\[ S2 = h^2 + k^2 - 3ah + \frac{3a^2}{4} \]
\[ \text{Length}^2 = h^2 + k^2 - 3ah + \frac{3a^2}{4} \]

3. For S3:
\[ S3 = h^2 + k^2 - 5ah + \frac{25a^2}{4} \]
\[ \text{Length}^2 = h^2 + k^2 - 5ah + \frac{25a^2}{4} \]

### Step 5: Set Up the Equation
According to the problem, the sum of the squares of the lengths of the tangents is equal to \(b^2\):
\[ (h^2 + k^2 - ah) + (h^2 + k^2 - 3ah + \frac{3a^2}{4}) + (h^2 + k^2 - 5ah + \frac{25a^2}{4}) = b^2 \]

Combining like terms:
\[ 3h^2 + 3k^2 - 9ah + 8a^2 = b^2 \]

### Step 6: Locus of Point T
To find the locus, we replace h with x and k with y:
\[ 3x^2 + 3y^2 - 9ax + 8a^2 - b^2 = 0 \]

### Final Result
The locus of point T is given by:
\[ 3x^2 + 3y^2 - 9ax + (8a^2 - b^2) = 0 \]