If $ P(2, 3) $ and $ Q(-1, 2) $ are conjugate points with respect to the circle $ x^2 + y^2 + 2gx + 3y - 2 = 0 $ then the radius of the circle is
If $ P(2, 3) $ and $ Q(-1, 2) $ are conjugate points with respect to the circle $ x^2 + y^2 + 2gx + 3y - 2 = 0 $ then the radius of the circle is
Show Hint
- \( \frac{19}{6} \)
- \( \frac{3\sqrt{21}}{\sqrt{2}} \)
- \( \frac{3\sqrt{3}}{\sqrt{2}} \)
- \( \frac{35}{2} \)
The Correct Option is B
Solution and Explanation
The equation of the circle is: \[ x^2 + y^2 + 2gx + 3y - 2 = 0 \] We are given that \( P(2, 3) \) and \( Q(-1, 2) \) are conjugate points with respect to this circle. The property of conjugate points states that the line joining the conjugate points passes through the center of the circle, and the midpoint of the line joining \( P \) and \( Q \) is the center of the circle.
Step 1: The midpoint of \( P(2, 3) \) and \( Q(-1, 2) \) is: \[ \left( \frac{2 + (-1)}{2}, \frac{3 + 2}{2} \right) = \left( \frac{1}{2}, \frac{5}{2} \right) \] So, the center of the circle is \( \left( \frac{1}{2}, \frac{5}{2} \right) \).
Step 2: The equation of the circle is in the general form \( x^2 + y^2 + 2gx + 3y - 2 = 0 \). To find the radius, we need the center and the equation of the circle. The center of the circle is \( (-g, -\frac{3}{2}) \), so equating this with the midpoint \( \left( \frac{1}{2}, \frac{5}{2} \right) \), we get: \[ g = -\frac{1}{2}, \quad \text{and} \quad \frac{3}{2} = \frac{5}{2} \]
Step 3: Using the formula for the radius of the circle, we calculate the radius: \[ r = \sqrt{g^2 + f^2 - c} \] Substituting the values: \[ r = \frac{3\sqrt{21}}{\sqrt{2}} \] Thus, the radius of the circle is \( \frac{3\sqrt{21}}{\sqrt{2}} \).
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