question_answer

The equation of the tangent to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]which makes a triangle of area \[{{a}^{2}}\] with the co-ordinate axes, is

A)            \[x\pm y=a\sqrt{2}\]               

B)            \[x\pm y=\pm a\sqrt{2}\]

C)            \[x\pm y=2a\]                          

D)            \[x+y=\pm 2a\]

Correct Answer: B

Solution :

           Let the tangent be of form \[\frac{x}{{{x}_{1}}}+\frac{y}{{{y}_{1}}}=1\] and area of \[\Delta \] formed by it with coordinate axes is                    \[\frac{1}{2}{{x}_{1}}{{y}_{1}}={{a}^{2}}\]                                                          ?.(i)                    Again, \[{{y}_{1}}x+{{x}_{1}}y-{{x}_{1}}{{y}_{1}}=0\]                    Applying conditions of tangency                    \[\left| \frac{-{{x}_{1}}{{y}_{1}}}{\sqrt{x_{1}^{2}+y_{1}^{2}}} \right|\ =a\] or \[(x_{1}^{2}+y_{1}^{2})=\frac{x_{1}^{2}y_{1}^{2}}{{{a}^{2}}}\]  ?.(ii)                    From (i) and (ii), we get\[{{x}_{1}},{{y}_{1}}\]; which gives equation of tangent as\[x\pm y=\pm a\sqrt{2}\].                    Trick: There may be 4 tangents (as in figure). As the lines \[x\pm y=\pm a\sqrt{2}\] make triangle of area ?a? in all four quadrants.