If the length of the tangents from P(1, 3) and Q (3, 7) to a circle are `sqrt2` units and `sqrt(18)` units respectively, then the length of the tangent from R(7, 15) to the same circle is

Options

A. `sqrt(98)` units
B. `sqrt(170)` units
C. `sqrt(50)` units
D. None of these

Answer

To solve the problem step by step, we will use the formula for the length of the tangent from a point to a circle. The length of the tangent \( L \) from a point \( (x_1, y_1) \) to a circle defined by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:

\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]

### Step 1: Set up the equations for points P and Q

For point \( P(1, 3) \) with tangent length \( \sqrt{2} \):

\[ L_P = \sqrt{1^2 + 3^2 + 2g(1) + 2f(3) + c} = \sqrt{2} \]

Squaring both sides:

\[ 1 + 9 + 2g + 6f + c = 2 \]

This simplifies to:

\[ 2g + 6f + c = 2 - 10 = -8 \quad \text{(Equation 1)} \]

For point \( Q(3, 7) \) with tangent length \( \sqrt{18} \):

\[ L_Q = \sqrt{3^2 + 7^2 + 2g(3) + 2f(7) + c} = \sqrt{18} \]

Squaring both sides:

\[ 9 + 49 + 6g + 14f + c = 18 \]

This simplifies to:

\[ 6g + 14f + c = 18 - 58 = -40 \quad \text{(Equation 2)} \]

### Step 2: Set up the equation for point R

For point \( R(7, 15) \) with tangent length \( L_R \):

\[ L_R = \sqrt{7^2 + 15^2 + 2g(7) + 2f(15) + c} \]

Squaring both sides gives us:

\[ 49 + 225 + 14g + 30f + c = L_R^2 \]

This simplifies to:

\[ 14g + 30f + c = L_R^2 - 274 \quad \text{(Equation 3)} \]

### Step 3: Solve the system of equations

Now we have three equations:

1. \( 2g + 6f + c = -8 \) (Equation 1)
2. \( 6g + 14f + c = -40 \) (Equation 2)
3. \( 14g + 30f + c = L_R^2 - 274 \) (Equation 3)

We can eliminate \( c \) by subtracting Equation 1 from Equation 2:

\[ (6g + 14f + c) - (2g + 6f + c) = -40 + 8 \]

This simplifies to:

\[ 4g + 8f = -32 \quad \Rightarrow \quad g + 2f = -8 \quad \text{(Equation 4)} \]

Next, we can eliminate \( c \) again by subtracting Equation 2 from Equation 3:

\[ (14g + 30f + c) - (6g + 14f + c) = L_R^2 - 274 + 40 \]

This simplifies to:

\[ 8g + 16f = L_R^2 - 234 \quad \Rightarrow \quad g + 2f = \frac{L_R^2 - 234}{8} \quad \text{(Equation 5)} \]

### Step 4: Equate Equations 4 and 5

From Equations 4 and 5, we have:

\[ -8 = \frac{L_R^2 - 234}{8} \]

Multiplying both sides by 8:

\[ -64 = L_R^2 - 234 \]

Thus:

\[ L_R^2 = 234 - 64 = 170 \]

### Step 5: Find \( L_R \)

Taking the square root:

\[ L_R = \sqrt{170} \]

### Final Answer

The length of the tangent from point \( R(7, 15) \) to the circle is \( \sqrt{170} \).

---