The straight line x + 2y = 1 meets the coordi | vedclass

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Question

Equation of circle $x\left( {x - 1} \right) + \left( {y - \frac{1}{2}} \right)y = 0$

${x^2} + {y^2} - x - \frac{y}{2} = 0$

Equation of tang

Vedclass · Accepted Answer

$\frac {\sqrt 5}{2}$

Vedclass · Answer

Equation of circle $x\left( {x - 1} ight) + \left( {y - \frac{1}{2}} ight)y = 0$

${x^2} + {y^2} - x - \frac{y}{2} = 0$

Equation of tangent at $\left( {0,0} ight)$

$x.0 + y.0 - \frac{{x + 0}}{2} - \frac{{y + 0}}{{2 	imes 2}} = 0$

$2x + y = 0\,\,\,\,\,\,\,\,.......\left( 1 ight)$

Sum of distance of $A$ and $B$ from line $(i)$ is

$\frac{2}{{\sqrt 5 }} + \frac{{\frac{1}{2}}}{{\sqrt 5 }} = \frac{5}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{2}$

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

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