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