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  • The equation of the circle with centre on the line and touching the lines

The equation of the circle with centre on the line \mathrm{2 x+y=0} and touching the lines \mathrm{4 x-3 y+10=0\: and \: 4 x-3 y-30=0} is

Option: 1

\mathrm{\quad(x-1)^{2}+(y+2)^{2}=16}


Option: 2

\mathrm{\quad(x-2)^{2}+(y+4)^{2}=16}


Option: 3

\mathrm{\quad(x+1)^{2}+(y-2)^{2}=16}


Option: 4

none of these.


Answers (1)

best_answer

Since the two lines are tangents to the given circle and they are parallel, we have

\text{Radius}=\frac{1}{2}\left|\frac{\mathrm{C}_{1}-\mathrm{C}_{2}}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}\right|=\frac{1}{2}\left|\frac{10+30}{\sqrt{16+9}}\right|=4

(we know that if \mathrm{a x+b y+c=0} and \mathrm{a x+b y+c^{\prime}= 0} are two parallel tangents to a circle, then equation of the line parallel to given lines and passing through  the centre is given by \mathrm{a x+ by +\frac{\mathrm{c}+\mathrm{c}^{\prime}}{2}=0}


Hence the centre of the circle lies on \mathrm{4 x-3 y-10=0}

Also the centre lies on \mathrm{ x+y=0}
Hence the coordinates of the centre are (1,-2).
\Rightarrow The equation of the circle is \mathrm{(x-1)^{2}+(y+2)^{2}=16}.

Hence (A) is the correct answer.

 

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chirag

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