The point of intersection of the tangents drawn at the ends of the chord joining the points $α$ and $β$ on the circle $x^{2} + y^{2} = a^{2}$ is

$(\frac{a sin \frac{α + β}{2}}{sin \frac{α - β}{2}}, \frac{a c o s \frac{α + β}{2}}{c o s \frac{α - β}{2}})$

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(\frac{a cos \frac{α + β}{2}}{cos \frac{α - β}{2}}, \frac{a sin \frac{α + β}{2}}{cos \frac{α - β}{2}})

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$(\frac{a cos \frac{α - β}{2}}{c o s \frac{α + β}{2}} \frac{a cos \frac{α - β}{2}}{sin \frac{α + β}{2}})$

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$(\frac{a cos \frac{α - β}{2}}{cos \frac{α + β}{2}}, \frac{a sin \frac{α - β}{2}}{sin \frac{α + β}{2}})$

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Solution

The correct option is B $(\frac{a cos \frac{α + β}{2}}{cos \frac{α - β}{2}}, \frac{a sin \frac{α + β}{2}}{cos \frac{α - β}{2}})$
\tan gent at $P (α) a n d P (B)$
$x cos α + y sin α = a - (1)$
$x cos β + y sin β = a - (2)$
Equation $(1) \to x cos α sin β + y sin β sin α = a sin β$
Equation $(2) \to x cos β sin α + y sin α sin β = a sin α$
$x (sin (β - α)) = a (sin β - sin α)$
$n = a x^{2} sin \frac{(\frac{β - α}{1}) cos (\frac{β + α}{2})}{2 sin (\frac{β - α}{2}) cos (\frac{β - α}{2})}$
$x = \frac{a cos (\frac{α + β}{2})}{cos (\frac{α - β}{2})}$
Then $y = a \frac{sin (\frac{α + β}{2})}{cos (\frac{α - β}{2})}$

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