Tangent $L_1 \equiv 3 x-4 y-8=0$ and the chord $L_2 \equiv x+y-1=0$ are at a distance of 2 and $\sqrt{2}$ units respectively from the centre of a circle S. $(h, k)$ is the centre of S such that $h^2+$ $\mathrm{k}^2=13$. If the midpoint of the chord $\mathrm{L}_2=0$ is $(\alpha, \beta)$ and the…

Question

Tangent $L_1 \equiv 3 x-4 y-8=0$ and the chord $L_2 \equiv x+y-1=0$ are at a distance of 2 and $\sqrt{2}$ units respectively from the centre of a circle S. $(h, k)$ is the centre of S such that $h^2+$ $\mathrm{k}^2=13$. If the midpoint of the chord $\mathrm{L}_2=0$ is $(\alpha, \beta)$ and the radius of the circle is $\mathrm{r}$, then $\alpha+\beta+\mathrm{r}=$

Aicharya · Accepted Answer

Radius of circle $=2=r$

$\because(\alpha, \beta)$ passes through $\mathrm{L}_2$. $\begin{aligned} & \therefore \alpha+\beta-1=0 \Rightarrow \alpha+\beta=1 \\ & \text { Now } \alpha+\beta+r=2+1=3 \end{aligned}$

Tangent $L_1 \equiv 3 x-4 y-8=0$ and the chord $L_2 \equiv x+y-1=0$ are at a distance of 2 and $\sqrt{2}$…

Solution