Given: The triangle \( OAB \) is an equilateral triangle, and the point \( P \) divides \( AB \) in the ratio \( 2:3 \).
\( \cos 60^\circ = \frac{OA^2 + AP^2 - OP^2}{2 \cdot OA \cdot AP} \).
\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{2 \cdot \lambda \cdot \frac{2\lambda}{5}} \).
\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{\frac{4\lambda^2}{5}} \).
\( \frac{2\lambda^2}{5} = \lambda^2 + \frac{4\lambda^2}{25} - OP^2 \).
\( OP^2 = \lambda^2 + \frac{4\lambda^2}{25} - \frac{2\lambda^2}{5} \).
\( OP^2 = \frac{25\lambda^2}{25} + \frac{4\lambda^2}{25} - \frac{10\lambda^2}{25} \).
\( OP^2 = \frac{19\lambda^2}{25} \).
\( OP = \sqrt{\frac{19\lambda^2}{25}} = \frac{\sqrt{19}}{5} \lambda \).
Final Answer: \( OP = \frac{\sqrt{19}}{5} \lambda \).
If the equation of the circle whose radius is 3 units and which touches internally the circle $$ x^2 + y^2 - 4x - 6y - 12 = 0 $$ at the point $(-1, -1)$ is $$ x^2 + y^2 + px + qy + r = 0, $$ then $p + q - r$ is:
Match List-I with List-II: List-I