Question

Let the locus of the point of intersection of the perpendicular tangents drawn to the circle \[ x^2 + y^2 + 6x - 4y - 12 = 0 \] be the circle \( S \). Then the equation of the tangent drawn to \( S \) which is perpendicular to the line \( 6x - 4y + k = 0 \) is:

• A. \( 4x + 6y \pm \sqrt{26} = 0 \)
• B. \( 2x + 3y \pm \sqrt{26} = 0 \)
• C. \( 2x + 3y \pm 5\sqrt{26} = 0 \)
• D. \( 4x + 6y \pm 5\sqrt{26} = 0 \)

Hint:

Use director circle concept for perpendicular tangents and apply standard form of tangent to a circle.

Answer 1

The Correct Option is C

Equation of the circle: \[ x^2 + y^2 + 6x - 4y - 12 = 0 \Rightarrow (x + 3)^2 + (y - 2)^2 = 25 \Rightarrow \text{Centre: } (-3, 2), \text{ Radius: } 5 \] The locus of intersection of perpendicular tangents to a circle is a new circle called the Director Circle: \[ x^2 + y^2 = 2r^2 = 50 \Rightarrow \text{Circle S: } x^2 + y^2 = 50 \] Now, find the tangent to this new circle \( x^2 + y^2 = 50 \) that is perpendicular to line \( 6x - 4y + k = 0 \) Slope of given line: \( \frac{3}{2} \Rightarrow \) slope of required line = \( -\frac{2}{3} \) Equation of tangent to a circle of form \( x^2 + y^2 = r^2 \) is: \[ lx + my = \pm r\sqrt{l^2 + m^2} \] Choose direction ratios \( l = 2, m = 3 \Rightarrow \text{Tangent: } 2x + 3y = \pm 5\sqrt{26} \Rightarrow \boxed{2x + 3y \pm 5\sqrt{26} = 0} \]