If the line $3 x-4 y=1$ touches the circle $(x-1)^2+(y+2)^2=4$ at $(\alpha, \beta)$, the values of $\alpha$…
If the line $3 x-4 y=1$ touches the circle $(x-1)^2+(y+2)^2=4$ at $(\alpha, \beta)$, the values of $\alpha$ and $\beta$ are
- $\alpha=\frac{1}{5}, \beta=-\frac{1}{10}$
- $\alpha=\frac{-1}{5}, \beta=-\frac{2}{5}$
- $\alpha=\frac{-2}{5}, \beta=\frac{-11}{20}$
- $\alpha=\frac{2}{5}, \beta=\frac{1}{20}$
Solution
$(\alpha, \beta)$ lies on the line $3 x-4 y=1$
$
\begin{array}{r}
3 \alpha-4 \beta=1 \\
\Rightarrow \quad \beta=\frac{(3 \alpha-1)}{4}
\end{array}
$
$(\alpha, \beta)$ also lie on the circle
$
\begin{aligned}
& \Rightarrow(\alpha-1)^2+(\beta+2)^2=4 \\
& \Rightarrow(\alpha-1)^2+\left(\frac{3 \alpha-1}{4}+2\right)^2=4 \\
& \Rightarrow 16(\alpha-1)^2+(3 \alpha+7)^2=64 \\
& \Rightarrow 16 \alpha^2+16-32 \alpha+9 \alpha^2+49+42 \alpha=64 \\
& \Rightarrow 25 \alpha^2+10 \alpha+1=0 \Rightarrow(5 \alpha+1)^2=0 \\
& \therefore \quad \alpha=\frac{-1}{5} \\
& \qquad 3\left(\frac{-1}{5}\right)-1 \\
& \text { and } \beta=\frac{-3-5}{4}=\frac{-8}{20}=\frac{-2}{5}
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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