Question

The equation of the pair of tangents drawn from the point \( (1, 1) \) to the circle \( x^2 + y^2 + 2x + 2y + 1 = 0 \) is:

• A. \( 3x^2 - 8xy + 3y^2 - 2x - 2y + 6 = 0 \)
• B. \( 11x^2 - 8xy + 11y^2 - 4x - 4y - 6 = 0 \)
• C. \( 3x^2 - 8xy + 3y^2 + 2x + 2y - 2 = 0 \)
• D. \( x^2 - 4xy + y^2 + x + y = 0 \)

Hint:

When solving for the equation of tangents from a point to a circle, complete the square for the circle’s equation and then use the formula for the tangents.

Answer 1

The Correct Option is C

The equation of the circle is given as: \[ x^2 + y^2 + 2x + 2y + 1 = 0 \] Step 1: Complete the square for the circle. First, complete the square for both \(x\) and \(y\) terms: \[ (x + 1)^2 + (y + 1)^2 = 1 \] So, the center of the circle is \( (-1, -1) \) and the radius is \( 1 \). Step 2: Use the formula for the equation of tangents. The formula for the equation of the pair of tangents from a point \( (x_1, y_1) \) to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is: \[ T = xx_1 + yy_1 + g(x_1 + x) + f(y_1 + y) + c = 0 \] Substitute the point \( (1, 1) \) and the center \( (-1, -1) \) into the formula to get the equation of the tangents. After simplification, the equation of the tangents is: \[ 3x^2 - 8xy + 3y^2 + 2x + 2y - 2 = 0 \]