Question

The equation of a tangent to the circle \(x^2 + y^2 + 2x - 12y - 132 = 0\) which is perpendicular to the line \(12x + 5y + k = 0\) is:

• A. \(5x - 12y + 92 = 0\)
• B. \(5x - 12y - 246 = 0\)
• C. \(5x - 12y - 169 = 0\)
• D. \(5x - 12y + 246 = 0\)

Hint:

Use slope-perpendicularity rule and point-to-line distance formula to find tangent line.

Answer 1

The Correct Option is D

The given line has slope \(m_1 = -\frac{12}{5}\), so perpendicular tangent must have slope \(m_2 = \frac{5}{12}\) Equation of line: \(5x - 12y + c = 0\) Find the value of \(c\) such that the line is tangent to the circle. Convert circle to standard form: Complete the square: \[ x^2 + 2x + y^2 - 12y = 132 \Rightarrow (x+1)^2 + (y - 6)^2 = 169 \] So center = \((-1, 6)\), radius = 13 Now, perpendicular distance from center to tangent: \[ \text{Distance} = \frac{|5(-1) - 12(6) + c|}{\sqrt{25 + 144}} = \frac{| -5 - 72 + c|}{13} = 13 \Rightarrow |-77 + c| = 169 \Rightarrow c = 246 \] So required tangent: \(5x - 12y + 246 = 0\)