Key Concepts and Formulas

• Section Formula: If a point $P$ divides the line segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$, then the coordinates of $P$ are given by $P\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
• Equation of a Circle: A circle with center $(h, k)$ and radius $r$ has the equation $(x-h)^2 + (y-k)^2 = r^2$. A circle with center at the origin has the equation $x^2 + y^2 = r^2$.
• Law of Cosines: In triangle $ABC$, $c^2 = a^2 + b^2 - 2ab\cos{C}$, where $a$, $b$, and $c$ are the side lengths opposite to angles $A$, $B$, and $C$ respectively.

Step-by-Step Solution

Step 1: Setting up the Coordinate System and Points

Let the circle be centered at the origin $O(0,0)$ with radius $\lambda$. Let the coordinates of point $A$ be $(\lambda \cos{\alpha}, \lambda \sin{\alpha})$ and the coordinates of point $B$ be $(\lambda \cos{\beta}, \lambda \sin{\beta})$. Since $A$ and $B$ lie on the circle, this parametrization is valid. We are given that $AB = \lambda$.

Step 2: Applying the Law of Cosines to Triangle AOB

Consider the triangle $AOB$. We have $OA = OB = \lambda$ and $AB = \lambda$. By the Law of Cosines, $AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos{\angle AOB}$ $\lambda^2 = \lambda^2 + \lambda^2 - 2\lambda^2 \cos{\angle AOB}$ $\lambda^2 = 2\lambda^2 - 2\lambda^2 \cos{\angle AOB}$ $2\lambda^2 \cos{\angle AOB} = \lambda^2$ $\cos{\angle AOB} = \frac{1}{2}$ Therefore, $\angle AOB = \frac{\pi}{3}$ or $60^\circ$. This means that $|\alpha - \beta| = \frac{\pi}{3}$. Without loss of generality, let $\beta = \alpha + \frac{\pi}{3}$.

Step 3: Finding the Coordinates of the Dividing Point P

Let $P(x,y)$ be the point that divides $AB$ in the ratio $2:3$. Using the section formula, we have: $x = \frac{2(\lambda \cos{\alpha}) + 3(\lambda \cos{\beta})}{2+3} = \frac{2\lambda \cos{\alpha} + 3\lambda \cos{(\alpha + \frac{\pi}{3})}}{5}$ $y = \frac{2(\lambda \sin{\alpha}) + 3(\lambda \sin{\beta})}{2+3} = \frac{2\lambda \sin{\alpha} + 3\lambda \sin{(\alpha + \frac{\pi}{3})}}{5}$ Simplifying the expressions for $x$ and $y$, we have: $x = \frac{\lambda}{5} \left(2\cos{\alpha} + 3\left(\cos{\alpha}\cos{\frac{\pi}{3}} - \sin{\alpha}\sin{\frac{\pi}{3}}\right)\right) = \frac{\lambda}{5} \left(2\cos{\alpha} + 3\left(\frac{1}{2}\cos{\alpha} - \frac{\sqrt{3}}{2}\sin{\alpha}\right)\right)$ $x = \frac{\lambda}{5} \left(2\cos{\alpha} + \frac{3}{2}\cos{\alpha} - \frac{3\sqrt{3}}{2}\sin{\alpha}\right) = \frac{\lambda}{5} \left(\frac{7}{2}\cos{\alpha} - \frac{3\sqrt{3}}{2}\sin{\alpha}\right)$ $y = \frac{\lambda}{5} \left(2\sin{\alpha} + 3\left(\sin{\alpha}\cos{\frac{\pi}{3}} + \cos{\alpha}\sin{\frac{\pi}{3}}\right)\right) = \frac{\lambda}{5} \left(2\sin{\alpha} + 3\left(\frac{1}{2}\sin{\alpha} + \frac{\sqrt{3}}{2}\cos{\alpha}\right)\right)$ $y = \frac{\lambda}{5} \left(2\sin{\alpha} + \frac{3}{2}\sin{\alpha} + \frac{3\sqrt{3}}{2}\cos{\alpha}\right) = \frac{\lambda}{5} \left(\frac{7}{2}\sin{\alpha} + \frac{3\sqrt{3}}{2}\cos{\alpha}\right)$

Step 4: Finding the Locus of P

Now we need to find the equation of the locus of $P$. To do this, we calculate $x^2 + y^2$: $x^2 + y^2 = \left(\frac{\lambda}{5}\right)^2 \left[\left(\frac{7}{2}\cos{\alpha} - \frac{3\sqrt{3}}{2}\sin{\alpha}\right)^2 + \left(\frac{7}{2}\sin{\alpha} + \frac{3\sqrt{3}}{2}\cos{\alpha}\right)^2\right]$ $x^2 + y^2 = \frac{\lambda^2}{25} \left[\frac{49}{4}\cos^2{\alpha} - \frac{42\sqrt{3}}{4}\cos{\alpha}\sin{\alpha} + \frac{27}{4}\sin^2{\alpha} + \frac{49}{4}\sin^2{\alpha} + \frac{42\sqrt{3}}{4}\sin{\alpha}\cos{\alpha} + \frac{27}{4}\cos^2{\alpha}\right]$ $x^2 + y^2 = \frac{\lambda^2}{25} \left[\frac{49}{4}(\cos^2{\alpha} + \sin^2{\alpha}) + \frac{27}{4}(\sin^2{\alpha} + \cos^2{\alpha})\right]$ $x^2 + y^2 = \frac{\lambda^2}{25} \left[\frac{49}{4} + \frac{27}{4}\right] = \frac{\lambda^2}{25} \left[\frac{76}{4}\right] = \frac{\lambda^2}{25}(19) = \frac{19}{25}\lambda^2$ Thus, $x^2 + y^2 = \left(\frac{\sqrt{19}}{5}\lambda\right)^2$.

Step 5: Identifying the Radius of the Locus

The locus of point $P$ is a circle centered at the origin with radius $\frac{\sqrt{19}}{5}\lambda$.

Common Mistakes & Tips

• Be careful with trigonometric identities and the section formula. Double-check your calculations.
• Choosing a convenient coordinate system (center of the circle at the origin) simplifies the calculations significantly.
• The condition $AB = \lambda$ is crucial and needs to be incorporated using the Law of Cosines.

Summary

We set up a coordinate system with the circle centered at the origin. Using the Law of Cosines and the given condition $AB=\lambda$, we found the angle subtended by the chord $AB$ at the center. Then, we applied the section formula to find the coordinates of the point $P$ that divides $AB$ in the ratio $2:3$. Finally, we determined the locus of $P$ to be a circle centered at the origin, and we calculated its radius to be $\frac{\sqrt{19}}{5}\lambda$.

Final Answer

The final answer is $\boxed{{\frac{\sqrt {19} } {5}}\lambda }$, which corresponds to option (D).