The point of intersection of the tangents drawn at the ends of the chord joining the points $\alpha$ and $\beta$ on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ is
- <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\sin\dfrac {\alpha+\beta}{2}}{\sin\dfrac {\alpha-\beta}{2}},\dfrac {\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha+\beta}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha-\beta}{2}})$</p>
- $(\displaystyle \dfrac{\mathrm{a}\cos\dfrac{\alpha+\beta}{2}}{\cos\dfrac{\alpha-\beta}{2}},\dfrac{\mathrm{a}\sin\dfrac{\alpha+\beta}{2}}{\cos\dfrac{\alpha-\beta}{2}})$
- <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}\dfrac {\alpha+\beta}{2}}\dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\sin\dfrac {\alpha+\beta}{2}})$</p>
- <p class="MsoNormal">$(\displaystyle \dfrac {\mathrm{a}\cos\dfrac {\alpha-\beta}{2}}{\cos\dfrac {\alpha+\beta}{2}},\dfrac {\mathrm{a}\sin\dfrac {\alpha-\beta}{2}}{\sin\dfrac {\alpha+\beta}{2}})$</p>
The tangent at a point with parameter angle t on x^2 + y^2 = a^2 is x cos t + y sin t = a. Solving the two tangent equations for t = alpha and t = beta gives the coordinates in option B.
The parametric coordinates for the points alpha and beta on the circle x squared plus y squared equals a squared are (a cos alpha, a sin alpha) and (a cos beta, a sin beta). The equation of the chord joining these two points is x cos((alpha plus beta) divided by 2) plus y sin((alpha plus beta) divided by 2) equals a cos((alpha minus beta) divided by 2). The point of intersection of the tangents at these points acts as the pole of this chord, so using the polar relationship for a circle gives the coordinates as (a cos((alpha plus beta) divided by 2) divided by cos((alpha minus beta) divided by 2), a sin((alpha plus beta) divided by 2) divided by cos((alpha minus beta) divided by 2)).