Question

Consider the lines $L: (k+7)x - (k-1)y - 4(k-5) = 0$ where k is a parameter and the circle $C: x^2 + y^2 + 4x + 12y - 60 = 0$ Statement-1: Every member of L intersects the circle 'C' at an angle of $90^\circ$ because Statement-2: Every member of L is tangent to the circle C.

• A. Statement-1 is true, statement-2 is true; statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true; statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (C)

Statement-1 is true, statement-2 is false.

Answer 2

The circle C has its center at $(-2, -6)$ and a radius of $R = \sqrt{2^2 + 6^2 - (-60)} = \sqrt{4 + 36 + 60} = \sqrt{100} = 10$. The family of lines L can be rewritten as $k(x - y - 4) + (7x + y + 20) = 0$. This indicates that all lines in L pass through the fixed point of intersection of $x - y - 4 = 0$ and $7x + y + 20 = 0$. Solving these equations yields $x = -2$ and $y = -6$. Thus, all lines in L pass through the center of the circle C.

Statement-2 is false because a line passing through the center of a circle cannot be tangent to it unless the radius is zero. The distance from the center to any line in L is 0, which is not equal to the radius 10.

Statement-1 is true because any line passing through the center of a circle intersects the circle at a $90^\circ$ angle. At the point of intersection, the radius lies along the line, and the tangent to the circle is perpendicular to the radius. Therefore, the angle between the line and the tangent is $90^\circ$.

Since Statement-1 is true and Statement-2 is false, option (C) is the correct answer.