1.

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

Answer»

`2sqrt5`
`(sqrt5)/(4)`
`4sqrt5`
`(sqrt5)/(2)`

Solution :According to given information, we have the following figure.

From figure, equation of circle (DIAMETER form) is
`(X-1)(x-0)+(y-0)((y-(1)/(2))=0`
`rArr x^(2) + y^(2) -x -(y)/(2)=0`
Equation of tangent at (0, 0) is `x+(y)+(2) = 0`
`[therefore" equation of tangent at "(x_(1),y_(1)) " is given by " T=0 " Here ", T = 0`
`rArr "xx"_(1)+yy_(1)-(1)/(2)(x+x_(1))-(1)/(4)(y+y_(1))=0]`
`rArr2x + y =0 `
Now, `AM=(|2.1+1.0|)/(sqrt5)=(2)/(sqrt5)`
`[therefore " distance of a point " P(x_(1), y_(1))`from a line
`ax+by + c=0 " is " (|ax_(1)+by_(1)+c|)/(sqrt(a^(2)+b^(2)))]`
and `BN = (|2.0+1((1)/(2))|)/(sqrt5)=(1)/(2sqrt5)`
`therefore AM+BN=(2)/(sqrt5)+(1)/(2sqrt5)=(4+1)/(2sqrt5)=(sqrt5)/(2)`


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