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 the perpendicular distances from A and B on the tangent to the circle at the origin is

Engineering

Mathematics

Tangent and Normal to Circle

Question

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

$\frac{\sqrt{5}}{4}$

$4 \sqrt{5}$

$2 \sqrt{5}$

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

$x^{2} + y^{2} - x - \frac{y}{2} = 0$

Equation of tangent at (0, 0)

$x . 0 + y . 0 - \frac{x + 0}{2} - \frac{y + 0}{2 \times 2} = 0$

2x + y = 0 …………(1)

Sum of distance of A and B form Line (i) is

$\frac{2}{\sqrt{5}} + \frac{\frac{1}{2}}{\sqrt{5}} = \frac{5}{2 \sqrt{5}} = \frac{\sqrt{5}}{2}$