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  • A circle through the point (2,8) touches the lines <img alt="\mathrm{4 x-3 y-24=0 \, \, and \, \, 4 x+3 y-42=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B4%20x-3%20y-24

A circle through the point (2,8) touches the lines \mathrm{4 x-3 y-24=0 \, \, and \, \, 4 x+3 y-42=0}. The x-coordinate of its centre is less than 8 . Its radius is

Option: 1

5


Option: 2

10


Option: 3

\frac{205}{3}


Option: 4

\frac{205}{6}


Answers (1)

best_answer

Since the circle touches the two lines, its centre lies on a bisector of angle between the lines:

\mathrm{ \begin{aligned} & \frac{4 x-3 y-24}{5}=\frac{ \pm(4 x+3 y-42)}{5} \\ & \Rightarrow y=3 \text { and } x=\frac{33}{4} \end{aligned} }

Let the centre be \mathrm{C(\alpha, 3).} The circle passes through (2,8). If r is the radius, then

\mathrm{ r^2=(\alpha-2)^2+(3-8)^2=(\alpha-2)^2+25 }                       ..(i)

The distance of $(\alpha, 3)$ from a line is r

\mathrm{ \begin{aligned} & \therefore \frac{4 \alpha-33}{5}=r \Rightarrow r^2=\frac{(4 \alpha-33)^2}{25} \, \, \, \, \, \, \, ...(ii)\\\\ & \text { (i), (ii) } \Rightarrow 25\left[\alpha^2-4 \alpha+29\right]=16 \alpha^2-264 \alpha+1089 \\\\ & \Rightarrow 9 \alpha^2+164 \alpha-364=0 . \end{aligned} }

(i), (ii) \mathrm{\Rightarrow 25\left[\alpha^2-4 \alpha+29\right]=16 \alpha^2-264 \alpha+1089}

\mathrm{\Rightarrow 9 \alpha^2+164 \alpha-364=0.}

Solving, \mathrm{\alpha=2, \frac{-182}{9}}

Now (ii) gives \mathrm{r=5, \frac{205}{9}.}

Posted by

Rakesh

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