The equation of the tangent to the circle `x^2+y^2=a^2,` which makes a triangle of area `a^2` with the coordinate axes, is `x+-y=asqrt(2)` (b) `x+-y=+-asqrt(2)` `x+-y=2a` (d) `x+y=+-2a`

Options

A. `x+- y= +-a`
B. `x+-y = +- a sqrt(2)`
C. `x+-y=3a`
D. `x+-y= +-2a`

Answer

2

Explanation

Let the tangent be of the form `(x)/(x_(1))+(y)/(y_(1))=1` and the area of triangle formed by it with the coordinate axes be
`(1)/(2) x_(1)y_(2)=a^(2) ` (1)
Using, the conditions of tangency , we get
`(| - x_(1)y_(1)|)/(sqrt(x_(1)^(2)+y_(1)^(2)))=a` or `x_(1)^(2)+y_(1)^(2)=(x_(1)^(2)y_(1)^(2))/(a^(2))` (2)
From (1) and (2) , we get the value of `x_(1)` and `y_(1)` , which gives the equation of tangent as `x+-y= +- a sqrt(2)`.