Question

The point diametrically opposite to the point \( P(1, 0) \) { on the circle} \[ x^2 + y^2 + 2x + 4y - 3 = 0 { is:} \]

• A. \( (3, -4) \)
• B. \( (-3, 4) \)
• C. \( (-3, -4) \)
• D. \( (3, 4) \)

Hint:

The diametrically opposite point on a circle is found by using the midpoint formula between the point and the center of the circle.

Answer 1

The Correct Option is C

Step 1: The given equation of the circle is: \[ x^2 + y^2 + 2x + 4y - 3 = 0. \] To rewrite this equation in standard form, we complete the square for both \( x \) and \( y \). 
Step 2: For \( x \), the coefficient of \( x \) is 2. Half of 2 is 1, and \( 1^2 = 1 \). Add and subtract 1 inside the equation. For \( y \), the coefficient of \( y \) is 4. Half of 4 is 2, and \( 2^2 = 4 \). Add and subtract 4 inside the equation: \[ (x^2 + 2x + 1) + (y^2 + 4y + 4) = 3 + 1 + 4. \] This simplifies to: \[ (x + 1)^2 + (y + 2)^2 = 8. \] Thus, the center of the circle is \( (-1, -2) \) and the radius is \( \sqrt{8} \). 
Step 3: The point diametrically opposite to \( P(1, 0) \) is on the straight line joining \( P \) and the center of the circle. The midpoint of \( P(1, 0) \) and the diametrically opposite point \( Q \) is the center \( (-1, -2) \). The midpoint formula is: \[ \left( \frac{1 + x_2}{2}, \frac{0 + y_2}{2} \right) = (-1, -2). \] Solving for \( x_2 \) and \( y_2 \): \[ \frac{1 + x_2}{2} = -1 \quad \Rightarrow \quad 1 + x_2 = -2 \quad \Rightarrow \quad x_2 = -3, \] \[ \frac{0 + y_2}{2} = -2 \quad \Rightarrow \quad y_2 = -4. \] Thus, the point diametrically opposite to \( P(1, 0) \) is \( (-3, -4) \).