Question

Equation of tangent to the circle \(x^2 + y^2 - 2x - 2y + 1 = 0\) perpendicular to \(y = x\) is given by

• A. \(x + y + 1 = 0\)
• B. \(x + y = 2 + \sqrt{3}\)
• C. \(x - y + 3 = 0\)
• D. \(x - y + 1 = 0\)

Hint:

For tangent: use distance from center = radius.

Answer 1

The Correct Option is B

Step 1: Find center and radius.
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \Rightarrow (x-1)^2 + (y-1)^2 = 1 \] Center \(C(1,1)\), radius \(r=1\)

Step 2: Slope of tangent.
Given line \(y=x\) has slope \(1\) Perpendicular slope \(=-1\) So tangent: \[ x + y + c = 0 \]

Step 3: Use distance formula.
Distance from center to tangent = radius: \[ \frac{|1+1+c|}{\sqrt{2}} = 1 \] \[ |2+c| = \sqrt{2} \Rightarrow c = -2 \pm \sqrt{2} \]

Step 4: Final equations.
\[ x + y = 2 \pm \sqrt{2} \] Matching options: \[ {x + y = 2 + \sqrt{2}} \]