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  • Find the radius (approximately) of smaller circle which touches the straight line <img alt="\mathrm{3x - y = 6}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B3x%20-%20y%20%3D%20

Find the radius (approximately) of smaller circle which touches the straight line \mathrm{3x - y = 6} at (1, -3) and also touches the line \mathrm{y = x. }

Option: 1

1


Option: 2

2


Option: 3

1.5


Option: 4

2.5


Answers (1)

best_answer

Let \mathrm{C}(\mathrm{h}, \mathrm{k}) be the center of the circle. Let \mathrm{AB}$ and $\mathrm{CD} be the lines represented by 3 \mathrm{x}-\mathrm{y}=6$ and $\mathrm{y}=\mathrm{x} respectively.
Clearly, the circle touches \mathrm{AB} at \mathrm{A}(1,-3)
Equation of line perpendicular to \mathrm{3 x-y=6 \: is \: x+3 y=\lambda} which passes through \mathrm{(1,-3)}
then \begin{aligned} & 1-9=\lambda \\ \end{aligned}

\mathrm{\therefore \lambda=-8}
\therefore perpendicular line is \mathrm{x}+3 \mathrm{y}+8=0

which passes through \mathrm{C}(\mathrm{h}, \mathrm{k})

then \mathrm{h+3 k+8=0}       ............(i)

Now center \mathrm{C}(\mathrm{h}, \mathrm{k}) \equiv \mathrm{C}(-3 \mathrm{k}-8, \mathrm{k})

radius \mathrm{CN}=\frac{|-3 \mathrm{k}-8-\mathrm{k}|}{\sqrt{1+1}}=\mathrm{CA}=\sqrt{(-3 \mathrm{k}-8-1)^2+(\mathrm{k}+3)^2}
\Rightarrow \quad \frac{4|\mathrm{k}+2|}{\sqrt{2}}=|\mathrm{k}+3| \sqrt{10}          ........(ii)
\Rightarrow \quad 4|\mathrm{k}+2|=2 \sqrt{5}|\mathrm{k}+3|
\Rightarrow \quad 2|\mathrm{k}+2|=\sqrt{5}|\mathrm{k}+3|
\therefore \quad \mathrm{k}+2= \pm \frac{\sqrt{5}}{2}(\mathrm{k}+3)
\therefore \quad \mathrm{k}=-7-2 \sqrt{5}$ or $\mathrm{k}=-7+2 \sqrt{5}
Since radius from (ii)

r=\frac{4|\mathrm{k}+2|}{\sqrt{2}}

\text { (radius) })_{\mathrm{at} k-7-2 \sqrt{5}}=\frac{4|-7-2 \sqrt{5}+2|}{\sqrt{2}}=10 \sqrt{2}+4 \sqrt{10}
=26.79
\text { (radius) })_{\mathrm{atk}-7+2 \sqrt{5}}=\frac{4|-7+2 \sqrt{5}+2|}{\sqrt{2}}=10 \sqrt{2}-4 \sqrt{10}
=1.5 \text { (nearly) }
Hence radius of smaller circle is 1.5 units. (nearly)

Hence option 2 is correct.

Posted by

jitender.kumar

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