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  • Let the centre of a circle be <img alt="(\alpha, \beta)" src="https://entrancecorner.oncodecogs.com/gif.latex?%28%5Calpha%2C%

Let the centre of a circle C be (\alpha, \beta) and its radius r<8. Let  3 x+4 y=24 and 3 x-4 y=32 be two tangents and 4 x+3 y=1 be a normal to C. Then (\alpha-\beta+r) is equal to

Option: 1

5


Option: 2

6


Option: 3

7


Option: 4

9


Answers (1)

best_answer

Centre lies on normal

\Rightarrow 4 \alpha+3 \beta=1\\ Distance \, \, of (\alpha, \beta) \, \, from L_1 \, \, and \, \, L_2 \, \, are \, \, equal\\\\ \left|\frac{3 \alpha+4 \beta-24}{5}\right|=\left|\frac{3 \alpha-4 \beta-32}{5}\right|

\begin{array}{ll} 3 \alpha+4 \beta-24=-3 \alpha+4 \beta+32 & 3 \alpha+4 \beta-24=-3 \alpha-4 \beta-32 \\ \Rightarrow 6 \alpha=56 & 8 \beta=-8 \\ \Rightarrow \alpha=\frac{28}{3}, \beta=\frac{-109}{3} & \beta=-1, \alpha=1 \\ \mathrm{r}=\sqrt{\left(\frac{28}{3}-4\right)^2+\left(\frac{-109}{3}+5\right)^2}>8 & \gamma=\sqrt{(4-1)^2+(-5+1)^2}=5 \\ \text { (reject) } & \alpha-\beta+\gamma=7 \end{array}

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HARSH KANKARIA

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