Let circle C be the image of x^2+y^2-2 x+4 y-4=0 in the line 2 x-3 y+5=0 and A be the poin…

Question

Let circle $C$ be the image of $x^2+y^2-2 x+4 y-4=0$ in the line $2 x-3 y+5=0$ and $A$ be the point on $C$ such that $O A$ is parallel to $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$

Accepted Answer

Centre $(1,-2), r=3$

Reflection of $(1,-2)$ about $2 x-3 y+5=0$

$$\begin{aligned}
& \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \
& x=-3, y=4
\end{aligned}$$

Equation of circle ' $C$ '

$C:(x+3)^2+(y-4)^2=9$

A.T.Q.

$$\begin{aligned}
& \ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r} \
& \mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r} \
& \theta=\frac{\pi}{3} \
& (\alpha+6)^2+(\beta-4)^2=27 \
& \frac{(\alpha+3)^2 \pm(\beta-4)^2=9}{(\alpha+6)^2-(\alpha+3)^2=18} \
& \Rightarrow 6 \alpha=-9 \
& \Rightarrow \alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right) \
& \therefore \beta-\sqrt{3} \alpha \
& \left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2} \
& =4
\end{aligned}$$

Solutions · 1