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The area of the triangle formed by the tangents from the points (h, k) to the circle \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\]and the line joining their points of contact is                                                   [MNR 1980]

A)            \[a\text{ }\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\]           

B)            \[a\text{ }\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{1/2}}}{{{h}^{2}}+{{k}^{2}}}\]

C)            \[\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\]         

D)            \[\frac{{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{1/2}}}{{{h}^{2}}+{{k}^{2}}}\]

Correct Answer: A

Solution :

           Equation of chord of contact AB is                    \[xh+yk={{a}^{2}}\]                     .....(i)                    \[OM=\]length of perpendicular from O(0, 0) on line (i)                          \[=\frac{{{a}^{2}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\]            \[\therefore \] \[AB=2AM=2\sqrt{O{{A}^{2}}-O{{M}^{2}}}=\frac{2a\sqrt{{{h}^{2}}+{{k}^{2}}-{{a}^{2}}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\]                    Also \[PM=\]length of perpendicular from \[P(h,\ k)\]to the line (i) is \[\frac{{{h}^{2}}+{{k}^{2}}-{{a}^{2}}}{\sqrt{{{h}^{2}}+{{k}^{2}}}}\]                    Therefore, the required area of triangle PAB                    \[=\frac{1}{2}.\ AB\ .\ PM=\frac{a{{({{h}^{2}}+{{k}^{2}}-{{a}^{2}})}^{3/2}}}{{{h}^{2}}+{{k}^{2}}}\].