Math, asked by StrongGirl, 8 months ago

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?

Answers

Answered by BrainlyConqueror0901
26

\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} }

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MisterIncredible: Awesome as always
Answered by Pikaachu
1

Step-by-step explanation:

  • Rewrite equation of straight line as :

l : x +  \frac{y}{ \frac{1}{2} }  = 1

  • which gives the points of intersection on axis as :

a : (1,0) \:, \: b : (0, \frac{1}{2} )

Now, equation of circle passing through points a, b and passing through origin is given as :

x(x - 1) + y(y -  \frac{1}{2} ) = 0

• Normal to the circle at origin ( line joining origin to centre ) is given by :

n:y = \frac{x}{2}

Hence, the tangent is given by :

T:y + 2x = 0

And so, sum of length of perpendicular drawn from A, B to given tangent is :

 \frac{ |0 + 2 \times 1| }{ \sqrt{ {2}^{2}  +  {1}^{2} } }  +  \frac{ | \frac{1}{2}  + 2 \times 0| }{ \sqrt{ {2}^{2}  +  {1}^{2} } }  =  \frac{ \sqrt{5} }{2}

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