Question

If the tangents \( x + y + k = 0 \) and \( x + ay + b = 0 \) drawn to the circle \( S : x^2 + y^2 + 2x - 2y + 1 = 0 \) are perpendicular to each other and \( k, b \) are both greater than 1, then find \( b - k \).

• A. \( \sqrt{2} \)
• B. 0
• C. 2
• D. \( \sqrt{2} \) \bigskip

Hint:

To solve for unknowns in problems involving perpendicular tangents, first express the tangents' slopes and use the condition that their product is \(-1\). Then, apply the distance formula from a point to a line.

Answer 1

The Correct Option is C

We are given the circle equation: \[ S: x^2 + y^2 + 2x - 2y + 1 = 0 \] and the tangent equations: \[ x + y + k = 0 \] \[ x + ay + b = 0 \] which are perpendicular to each other. We need to find \( b - k \) given that \( k, b>1 \). --- Step 1: Convert Circle Equation to Standard Form The given circle equation: \[ x^2 + y^2 + 2x - 2y + 1 = 0 \] We complete the square: \[ (x^2 + 2x) + (y^2 - 2y) + 1 = 0 \] \[ (x+1)^2 - 1 + (y-1)^2 - 1 + 1 = 0 \] \[ (x+1)^2 + (y-1)^2 = 2 \] Thus, the center of the circle is \( (-1,1) \) and the radius is \( \sqrt{2} \). --- Step 2: Find Slopes of Tangents The given tangent equations can be rewritten in slope-intercept form: 1. \( x + y + k = 0 \Rightarrow y = -x - k \) - Slope: \( m_1 = -1 \) 2. \( x + ay + b = 0 \Rightarrow y = -\frac{1}{a} x - \frac{b}{a} \) - Slope: \( m_2 = -\frac{1}{a} \) Since the tangents are perpendicular, their slopes satisfy: \[ m_1 \cdot m_2 = -1 \] \[ (-1) \times \left(-\frac{1}{a}\right) = -1 \] \[ \frac{1}{a} = -1 \] \[ a = -1 \] --- Step 3: Use the Tangent Distance Formula For a line \( Ax + By + C = 0 \) to be tangent to the circle \( (x+1)^2 + (y-1)^2 = 2 \), the perpendicular distance from the center \( (-1,1) \) to the line must be equal to the radius \( \sqrt{2} \). Using the perpendicular distance formula: \[ \frac{|A(-1) + B(1) + C|}{\sqrt{A^2 + B^2}} = \sqrt{2} \] For \( x + y + k = 0 \): \[ \frac{|-1 + 1 + k|}{\sqrt{1^2 + 1^2}} = \sqrt{2} \] \[ \frac{|k|}{\sqrt{2}} = \sqrt{2} \] \[ |k| = 2 \] Since \( k>1 \), we take \( k = 2 \). For \( x - y + b = 0 \) (substituting \( a = -1 \)): \[ \frac{|-1 -1 + b|}{\sqrt{1^2 + (-1)^2}} = \sqrt{2} \] \[ \frac{|b - 2|}{\sqrt{2}} = \sqrt{2} \] \[ |b - 2| = 2 \] Since \( b>1 \), we take \( b = 4 \). --- Step 4: Compute \( b - k \) \[ b - k = 4 - 2 = 2 \] --- Final Answer: \[ \boxed{2} \] \bigskip