MathematicsMedium127×since 2002Q3127If a line, y = mx + c is a tangent to the circle, (x – 3)² + y² = 1 and it is perpendicular to a line L₁, where L₁ is the tangent to the circle, x² + y² = 1 at the point (12,12)\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)(21,21), then :Ac² + 6c + 7 = 0Bc² - 7c + 6 = 0Cc² – 6c + 7 = 0Dc² + 7c + 6 = 0Check answerSkip