(a)OR(b)

A circle centered at $\displaystyle (2,1)$ passes through the points A$\displaystyle (5, 6)$ and B(-$\displaystyle 3$, K). Find the value(s) of K. Hence find length of chord AB.

Let centre be $\displaystyle \mathrm{O}(2,1) \Rightarrow \mathrm{OA}=\mathrm{OB}$ \[\begin{array}{l} \sqrt{(5-2)^{2}+(6-1)^{2}}=\sqrt{(-3-2)^{2}+(\mathrm{K}-1)^{2}} \\ \Rightarrow 9=(\mathrm{K}-1)^{2} \\ \Rightarrow \mathrm{~K}=-2,4 \end{array} \]For $\displaystyle \mathrm{K}=-2, \mathrm{AB}=\sqrt{128}$ or $\displaystyle 8 \sqrt{2}$ For $\displaystyle \mathrm{K}=4, \mathrm{AB}=\sqrt{68}$ or $\displaystyle 2 \sqrt{17}$Prove that the point P dividing the line segment joining the points $\displaystyle \mathrm{A}(-1,7)$ and $\displaystyle \mathrm{B}(4,-3)$ in the ratio $\displaystyle 3: 2$, lies on the line $\displaystyle x-3 \mathrm{y}=-1$. Also find length of PA and PB.X_Mathematics_041_30/$\displaystyle 5$/1_2025-$\displaystyle 26$ $\displaystyle \mathrm{AP}: \mathrm{PB}=3: 2$ Coordinates of $\displaystyle \mathrm{P}=\left(\frac{12-2}{5}, \frac{-9+14}{5}\right)=(2,1)$ Substituting $\displaystyle \mathrm{x}=2$ and $\displaystyle \mathrm{y}=1$ in the given equation L.H.S. $\displaystyle =x-3 y$ \[\begin{array}{l} =2-3(1) \\ =-1=\mathrm{R} . \mathrm{H} . \mathrm{S} . \end{array} \] $\displaystyle \therefore \mathrm{P}$ lies on the given line \[\begin{array}{l} \mathrm{PA}=\sqrt{(2+1)^{2}+(1-7)^{2}}=\sqrt{45} \text { or } 3 \sqrt{5} \\ \mathrm{~PB}=\sqrt{(2-4)^{2}+(1+3)^{2}}=\sqrt{20} \text { or } 2 \sqrt{5} \end{array} \]