Get Answers to all your Questions

header-bg qa

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 origing is:

  • Option 1)

     

    \frac{\sqrt5}{2}

  • Option 2)

     

    2\sqrt5

  • Option 3)

     

    4\sqrt5

     

  • Option 4)

     

    \frac{\sqrt5}{4}

Answers (1)

best_answer

 

Perpendicular distance of a point from a line -

\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}

 

 

- wherein

\rho  is the distance from the line ax+by+c=0 .

 

 

Equation of tangent -

xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0
 

- wherein

Tangent to circle

x^{2}+y^{2}+2gx+2fy+c=0  at  (x_{1},y_{1})

Equation of circle is

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

Equation of tangent at circle at \left ( 0,0 \right ) is

-x-\frac{y}{2}=0

\Rightarrow 2x+y=0\cdots \cdots \cdots \left ( 1 \right )

distance of \left ( 0,\frac{1}{2} \right )     from (1)  is  \frac{1}{2\sqrt{5}}

distance of \left ( 1,0 \right )        from (1) is  \frac{2}{\sqrt{5}}

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

 

 


Option 1)

 

\frac{\sqrt5}{2}

Option 2)

 

2\sqrt5

Option 3)

 

4\sqrt5

 

Option 4)

 

\frac{\sqrt5}{4}

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE