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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 co-ordinate axes, is
$x \pm y = a\sqrt 2 $
$x \pm y = \pm a\sqrt 2 $
$x \pm y = 2a$
$x + y = \pm 2a$
Solution
(b) 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 $(x_1^2 + y_1^2) = \frac{{x_1^2y_1^2}}{{{a^2}}}$….$(ii)$
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.
