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The equation of the circle with centre on the line $\mathrm{2x + y =0}$ and touching the lines

$\mathrm{4x - 3y + 10 = 0 }$ and $\mathrm{4x-3y-30 = 0}$ is $\mathrm{(x-a)^2+(y+2)^2=\lambda^2}$ , where $\mathrm{\lambda}$ and $\mathrm{\alpha }$ are
Option: 1
$\mathrm{ 2,1}$

Option: 2
$\mathrm{2,2}$

Option: 3
$\mathrm{2,-2}$

Option: 4
none of these

Answers (1)

best_answer

As tangents are parallel, it implies that diameter $\mathrm{=\left|\frac{30-10}{\sqrt{16+9}}\right|=4}$

Radius = 2 If $\mathrm{ ax + by + c = 0 }$ and $\mathrm{ a x+b y+c^{\prime}}$ are two parallel tangent lines, then a line parallel to them through the centre of circle is; $\mathrm{a x+b y+\frac{c+c^{\prime}}{2}=0 }$

$\mathrm{\left(A B=\left|\frac{c^{\prime}-c}{2}\right|, B C=\left|\frac{c-c^{\prime}}{2}\right|\right) }$

Hence, the centre of the circle $\mathrm{(\alpha, \beta) }$ lies on $\mathrm{4 x-3 y+\frac{10+(-30)}{2}=0 }$

$\mathrm{\Rightarrow 4 \alpha-3 \beta-10=0 }$

And, also on $\mathrm{2 x+y=0 \Rightarrow 2 \alpha+\beta=0 }$

On solving, $\mathrm{\alpha=1, \beta=-2 }$

$\mathrm{\therefore }$ The circle is $\mathrm{(x-1)^2+(y+2)^2=4 }$

Posted by

Deependra Verma

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