Question

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
A. $ x\pm y=\pm a $
B. $ x\pm y=\pm a\sqrt{2} $
C. $ x\pm y=3a $
D. $ x\pm y=\pm 2a $

Answer 1

Hint: We first try to assume the intercepts of the tangent on the axes. From that, we find the area of the triangle and also the equation of the tangent. Then we use the distance formula to form the radius as the perpendicular distance. We get two equations for the variables. We solve them to find the equation of the tangent.

Complete step by step answer:
The equation of the tangent to the circle $ {{x}^{2}}+{{y}^{2}}={{a}^{2}} $ makes a triangle of area $ {{a}^{2}} $ with the coordinate axes.

The tangent to the circle at point B intersects the axes at point C and D. let’s assume the coordinates of the points are $ C\equiv \left( 0,\alpha \right) $ and $ D\equiv \left( \beta ,0 \right) $ .
 $ \Delta COD $ being a right-angle triangle, we can find the area as \[\dfrac{1}{2}\times \alpha \times \beta =\dfrac{\alpha \beta }{2}\].
We have been given that the area is $ {{a}^{2}} $ , which means \[\dfrac{\alpha \beta }{2}={{a}^{2}}\].
We also know that the equation of a line with axes intercepts of $ \alpha $ and $ \beta $ for Y and X axes will be $ \dfrac{x}{\beta }+\dfrac{y}{\alpha }=1 $ .
For the given circle $ {{x}^{2}}+{{y}^{2}}={{a}^{2}} $ , the centre is $ \left( 0,0 \right) $ and the radius is a unit.
If we find the distance of the tangent line $ \dfrac{x}{\beta }+\dfrac{y}{\alpha }=1 $ from the centre $ \left( 0,0 \right) $ , then we will get the radius length.
We know that the distance from a point $ \left( m,n \right) $ to the line $ px+qy+c=0 $ is $ \left| \dfrac{pm+qn+c}{\sqrt{{{p}^{2}}+{{q}^{2}}}} \right| $ .
So, $ a=\left| \dfrac{-1}{\sqrt{\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}}} \right| $ which gives $ \dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}=\dfrac{1}{{{a}^{2}}} $ .
We have two equations \[\dfrac{\alpha \beta }{2}={{a}^{2}}\] and $ \dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}=\dfrac{1}{{{a}^{2}}} $ . We solve them.
So, $ \dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}=\dfrac{2}{a\beta } $ . We got this replacing value of a. This gives
\[\begin{align}
  & \Rightarrow \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}=\dfrac{2}{a\beta } \\
 & \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}-2a\beta =0 \\
 & \Rightarrow {{\left( \alpha -\beta \right)}^{2}}=0 \\
 & \Rightarrow \alpha =\beta \\
\end{align}\]
We place the value in \[\dfrac{\alpha \beta }{2}={{a}^{2}}\] and get \[\alpha =\beta =\pm \sqrt{2}a\].
Therefore, the equation of the tangent becomes $ \dfrac{x}{\pm \sqrt{2}a}+\dfrac{y}{\pm \sqrt{2}a}=1 $ which gives $ x\pm y=\pm a\sqrt{2} $ . The correct option is B.

Note:
We also could have taken the direct tangent theorem of a circle. For the general equation of circle $ {{x}^{2}}+{{y}^{2}}={{a}^{2}} $ , the equation of the tangent of the circle at point $ \left( k,l \right) $ is $ kx+yl={{a}^{2}} $ . From the equation, we would have found the intercepts and proceed the same way to find values of k and l.