Question

If the inverse point of the point \( (-1, 1) \) with respect to the circle \( x^2 + y^2 - 2x + 2y - 1 = 0 \) is \( (p, q) \), then \( p^2 + q^2 = \) 

• A. \( \frac{1}{16} \)
• B. \( \frac{1}{8} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{2} \)

Hint:

To find the inverse of a point with respect to a circle, use the inverse point formula and ensure to complete the square to find the center and radius of the circle.

Answer 1

The Correct Option is B

We are given the equation of the circle as: \[ x^2 + y^2 - 2x + 2y - 1 = 0 \] and the point \( (-1, 1) \). We are required to find the value of \( p^2 + q^2 \), where \( (p, q) \) is the inverse point of \( (-1, 1) \) with respect to the given circle.
Step 1: First, we find the center and radius of the given circle. We complete the square for both \( x \) and \( y \) in the equation of the circle.
The given equation is: \[ x^2 - 2x + y^2 + 2y = 1 \] Completing the square for \( x \) and \( y \): \[ (x - 1)^2 + (y + 1)^2 = 3 \] Thus, the center of the circle is \( (1, -1) \) and the radius is \( \sqrt{3} \).
Step 2: Now, the formula for the inverse of a point \( (x_1, y_1) \) with respect to a circle with center \( (h, k) \) and radius \( r \) is given by: \[ (x_1, y_1) \rightarrow \left( \frac{r^2(x_1 - h)}{(x_1 - h)^2 + (y_1 - k)^2} + h, \frac{r^2(y_1 - k)}{(x_1 - h)^2 + (y_1 - k)^2} + k \right) \] For the given point \( (-1, 1) \), the center of the circle is \( (1, -1) \), and the radius is \( \sqrt{3} \). Substituting into the formula: \[ p = \frac{3(-1 - 1)}{(-1 - 1)^2 + (1 + 1)^2} + 1 = \frac{3(-2)}{4 + 4} + 1 = \frac{-6}{8} + 1 = -\frac{3}{4} + 1 = \frac{1}{4} \] \[ q = \frac{3(1 + 1)}{(-1 - 1)^2 + (1 + 1)^2} - 1 = \frac{3(2)}{4 + 4} - 1 = \frac{6}{8} - 1 = \frac{3}{4} - 1 = -\frac{1}{4} \] Step 3: Now, we calculate \( p^2 + q^2 \): \[ p^2 + q^2 = \left( \frac{1}{4} \right)^2 + \left( -\frac{1}{4} \right)^2 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8} \] Thus, the value of \( p^2 + q^2 \) is \( \frac{1}{8} \).