Question

If A and B are the centres of similitude with respect to the circles \( x^2 + y^2 - 14x + 6y + 33 = 0 \) and \( x^2 + y^2 + 30x - 2y + 1 = 0 \), then midpoint of \( AB \) is:

• A. \( \left( \frac{7}{3}, \frac{4}{5} \right) \)
• B. \( \left( \frac{3}{2}, \frac{1}{5} \right) \)
• C. \( \left( \frac{39}{2}, \frac{-7}{4} \right) \)
• D. \( \left( \frac{39}{4}, \frac{-7}{2} \right) \)

Hint:

For midpoint, use \( M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \) formula.

Answer 1

The Correct Option is D

Step 1: Identify the Centres and Radii of the Given Circles The given circles are: \[ x^2 + y^2 - 14x + 6y + 33 = 0 \] \[ x^2 + y^2 + 30x - 2y + 1 = 0 \] Step 2: Complete the Square For the first circle: \[ x^2 - 14x + y^2 + 6y + 33 = 0 \] Completing the square: \[ (x - 7)^2 - 49 + (y + 3)^2 - 9 + 33 = 0 \] \[ (x - 7)^2 + (y + 3)^2 - 25 = 0 \] \[ (x - 7)^2 + (y + 3)^2 = 25 \] Thus, the center is \( (7, -3) \) and radius \( R_1 = 5 \). For the second circle: \[ x^2 + 30x + y^2 - 2y + 1 = 0 \] Completing the square: \[ (x + 15)^2 - 225 + (y - 1)^2 - 1 + 1 = 0 \] \[ (x + 15)^2 + (y - 1)^2 = 225 \] Thus, the center is \( (-15, 1) \) and radius \( R_2 = 15 \). Step 3: Centres of Similitude The centres of similitude are given by the section formula: \[ \mathbf{C_1} = \frac{R_2 \mathbf{O_1} + R_1 \mathbf{O_2}}{R_2 + R_1} \] \[ \mathbf{C_2} = \frac{R_2 \mathbf{O_1} - R_1 \mathbf{O_2}}{R_2 - R_1} \] Using the first formula: \[ \mathbf{C_1} = \frac{15(7, -3) + 5(-15, 1)}{15 + 5} \] \[ \mathbf{C_1} = \frac{(105, -45) + (-75, 5)}{20} \] \[ \mathbf{C_1} = \frac{(30, -40)}{20} = (1.5, -2) \] Using the second formula: \[ \mathbf{C_2} = \frac{15(7, -3) - 5(-15, 1)}{15 - 5} \] \[ \mathbf{C_2} = \frac{(105, -45) + (75, -5)}{10} \] \[ \mathbf{C_2} = \frac{(180, -50)}{10} = (18, -5) \] Step 4: Midpoint of \( AB \) The midpoint of \( AB \) is: \[ \text{Midpoint} = \left( \frac{1.5 + 18}{2}, \frac{-2 + (-5)}{2} \right) \] \[ = \left( \frac{19.5}{2}, \frac{-7}{2} \right) \] \[ = \left( \frac{39}{4}, \frac{-7}{2} \right) \] Step 5: Final Answer 

\[Correct Answer: (4) \ \left( \frac{39}{4}, \frac{-7}{2} \right)\]