Question

straight line x+2y=1 cuts coordinate axes at points A &B. A circle passes through points A, B origin. Then sum of the length of the perpendicular drawn from A&B on tangent at origin of given circle is?
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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{AC+BD=\frac{\sqrt{5}}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Straight \: line \to x + 2y =0 \\ \\ \tt: \implies Circle \: passing \: through \: points \: a,b \: and \: orgin \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Sum \: of \: length \: of \: perpendicular \:draw \: from \: point \: A \: and \: B = ?$

• According to given question :

$\bold{For \:coordinate \: of \: point \: A } \\ \tt: \implies x + 2y = 1 \\ \\ \tt \circ \: put \: y = 0 \\ \\ \tt: \implies x = 1 \\ \\ \bold{Similarly \: for \: point \: B : } \\ \tt: \implies x + 2y = 1 \\ \\ \tt \circ \: put \: x = 0 \\ \\ \tt: \implies y = \frac{1}{2} \\ \\ \bold{as \: we \: know \: that} \\ \tt: \implies (x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0 \\ \\ \tt \circ \: (x_{1}, y_{1}) = (0,0) \\ \\ \tt \circ \: (x_{2},y_{2}) = (1, \frac{1}{2} ) \\ \\ \tt: \implies (x - 0)(x - 1) + (y - 0)(y - \frac{1}{2}) = 0 \\ \\ \tt: \implies {x}^{2} - x + {y}^{2} - \frac{y}{2} =0 \\ \\ \green{\tt: \implies {x}^{2} + {y}^{2} - x - \frac{y}{2} = 0 \to \: eqn \: of \: circle}$

$\bold{For \: distance \:AC: } \\ \tt: \implies AC= \frac{2 \times 1 + 1 \times 0}{ \sqrt{ {2}^{2} + {1}^{2} } } \\ \\ \tt: \implies AC= \frac{2 }{ \sqrt{5} } - - - - - (1) \\ \\ \bold{For \: distance \: BD : } \\ \tt: \implies BD = \frac{2 \times 0 + 1 \times \frac{1}{2} }{ \sqrt{ {2}^{2} + {1}^{2} } } \\ \\ \tt: \implies BD = \frac{1}{2 \sqrt{5} } - - - - - (2)\\ \\ \bold{For \: sum \: of \: AC +BD} \\ \tt: \implies AC + BD = \frac{2}{\sqrt{5} } + \frac{1}{2 \sqrt{5} } \\ \\ \tt: \implies AC+ BD= \frac{4 + 1}{2 \sqrt{5} } \\ \\ \green{\tt: \implies AC+ BD = \frac{ \sqrt{5} }{2} } \\ \\ \green{ \tt \therefore Sum \: of \: perpendicular \: distance \: from \: A \:and \: B\: is \: \frac{ \sqrt{5} }{2} }$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{AC+BD=\frac{\sqrt{5}}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Straight \: line \to x + 2y =0 \\ \\ \tt: \implies Circle \: passing \: through \: points \: a,b \: and \: orgin \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Sum \: of \: length \: of \: perpendicular \:draw \: from \: point \: A \: and \: B = ?$

• According to given question :

$\bold{For \:coordinate \: of \: point \: A } \\ \tt: \implies x + 2y = 1 \\ \\ \tt \circ \: put \: y = 0 \\ \\ \tt: \implies x = 1 \\ \\ \bold{Similarly \: for \: point \: B : } \\ \tt: \implies x + 2y = 1 \\ \\ \tt \circ \: put \: x = 0 \\ \\ \tt: \implies y = \frac{1}{2} \\ \\ \bold{as \: we \: know \: that} \\ \tt: \implies (x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0 \\ \\ \tt \circ \: (x_{1}, y_{1}) = (0,0) \\ \\ \tt \circ \: (x_{2},y_{2}) = (1, \frac{1}{2} ) \\ \\ \tt: \implies (x - 0)(x - 1) + (y - 0)(y - \frac{1}{2}) = 0 \\ \\ \tt: \implies {x}^{2} - x + {y}^{2} - \frac{y}{2} =0 \\ \\ \green{\tt: \implies {x}^{2} + {y}^{2} - x - \frac{y}{2} = 0 \to \: eqn \: of \: circle}$

$\bold{For \: distance \:AC: } \\ \tt: \implies AC= \frac{2 \times 1 + 1 \times 0}{ \sqrt{ {2}^{2} + {1}^{2} } } \\ \\ \tt: \implies AC= \frac{2 }{ \sqrt{5} } - - - - - (1) \\ \\ \bold{For \: distance \: BD : } \\ \tt: \implies BD = \frac{2 \times 0 + 1 \times \frac{1}{2} }{ \sqrt{ {2}^{2} + {1}^{2} } } \\ \\ \tt: \implies BD = \frac{1}{2 \sqrt{5} } - - - - - (2)\\ \\ \bold{For \: sum \: of \: AC +BD} \\ \tt: \implies AC + BD = \frac{2}{\sqrt{5} } + \frac{1}{2 \sqrt{5} } \\ \\ \tt: \implies AC+ BD= \frac{4 + 1}{2 \sqrt{5} } \\ \\ \green{\tt: \implies AC+ BD = \frac{ \sqrt{5} }{2} } \\ \\ \green{ \tt \therefore Sum \: of \: perpendicular \: distance \: from \: A \:and \: B\: is \: \frac{ \sqrt{5} }{2} }$