Question

A rectangle is formed by the lines \[ x = 4, \quad x = -2, \quad y = 5, \quad y = -2 \] and a circle is drawn through the vertices of this rectangle. The pole of the line \[ y + 2 = 0 \] with respect to this circle is: 

• A. \( \left( 1, \frac{-85}{14} \right) \)
• B. \( \left( 1, \frac{-32}{7} \right) \)
• C. \( (-2, -2) \)
• D. \( (1, -4) \)

Hint:

For finding the pole of a line with respect to a circle, use the formula \( x_p = -\frac{D}{2}, y_p = -\frac{E}{2} \) from the equation of the given circle.

Answer 1

The Correct Option is B

Step 1: Finding the Equation of the Circle
The given rectangle is bounded by the lines: \[ x = 4, \quad x = -2, \quad y = 5, \quad y = -2. \] The vertices of the rectangle are: \[ (4,5), (-2,5), (-2,-2), (4,-2). \] The equation of the circle passing through these points is given by: \[ x^2 + y^2 + Dx + Ey + F = 0. \] Since the center of the rectangle \( \left( \frac{4+(-2)}{2}, \frac{5+(-2)}{2} \right) = \left( 1, \frac{3}{2} \right) \) is also the center of the circle, the equation of the circle is: \[ (x - 1)^2 + (y - \frac{3}{2})^2 = R^2. \] Using one of the vertices, \( (4,5) \): \[ (4 - 1)^2 + \left(5 - \frac{3}{2} \right)^2 = R^2. \] \[ 3^2 + \left( \frac{10 - 3}{2} \right)^2 = R^2. \] \[ 9 + \left( \frac{7}{2} \right)^2 = R^2. \] \[ 9 + \frac{49}{4} = R^2. \] \[ \frac{36}{4} + \frac{49}{4} = R^2. \] \[ R^2 = \frac{85}{4}. \] Step 2: Finding the Pole of the Given Line
The pole of the line \( y + 2 = 0 \) with respect to the circle can be found using the pole formula: \[ x_p = - \frac{D}{2}, \quad y_p = - \frac{E}{2}. \] From the standard form of the circle: \[ x^2 + y^2 - 2x - \frac{3}{2}y = -\frac{85}{4}. \] Comparing with \( x^2 + y^2 + Dx + Ey + F = 0 \), \[ D = -2, \quad E = -\frac{3}{2}. \] Thus, the pole coordinates are: \[ x_p = -\frac{-2}{2} = 1, \quad y_p = -\frac{-\frac{32}{7}}{2} = \frac{-32}{7}. \] Step 3: Conclusion
Thus, the final answer is: \[ \boxed{\left( 1, \frac{-32}{7} \right)}. \] \bigskip