The straight line $x+2 y=1$ meets the coordinate axes at $A$ and $\mathrm{B}$. A circle is drawn through $\mathrm{A}, \mathrm{B}$ and the origin. Then the sum of perpendicular distances from $\mathrm{A}$ and $\mathrm{B}$ on the tangent to the circle at the origin is:

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Aicharya · Accepted Answer

Let equation of circle be $x^{2}+y^{2}+2 g x+2 f y=0$ As length of intercept on $x$ axis is $1=2 \sqrt{g^{2}-c}$ $\Rightarrow|g|=\frac{1}{2}$ length of intercept on $y$ -axis $=\frac{1}{2}=2 \sqrt{f^{2}-c}$ $\Rightarrow|f|=\frac{1}{4}$ Equation of circle that passes through given points is $x^{2}+y^{2}-x-\frac{y}{2}=0$ Tangent at (0,0) is, $(y-0)=\left(\frac{d y}{d x}\right)_{(0,0)}^{(x-0)} \cdot(\mathrm{x}-0)$ $\Rightarrow 2 x+y=0$ Perpendicular distance from $B(1,0)$ on the tangent to the circle $=\frac{\frac{1}{2}}{\sqrt{5}}$ Perpendicular distance from $B\left(0, \frac{1}{2}\right)$ on the tangent to the circle $=\frac{2}{\sqrt{5}}$ Sum of perpendicular distance $=\frac{\frac{1}{2}+2}{\sqrt{5}}=\frac{\sqrt{5}}{2}$.

The straight line $x+2 y=1$ meets the coordinate axes at $A$ and $\mathrm{B}$. A circle is drawn through…

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