If a tangent to the ellipse x ^ 2 + 4 y ^ 2 = 4 meets the tangents...

We have, $\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{1} = 1$

Equation of tangent is \left(cos\theta\right)x + \left(2sin\theta\right)y = 2$</p><p>$B\left(- 2, \dfrac{1 + cos\theta}{sin\theta}\right), C\left(2, \dfrac{1 - cos\theta}{sin\theta}\right)$</p><p>$≡ B\left(- 2, cot\dfrac{\theta}{2}\right), C\left(2, tan\dfrac{\theta}{2}\right)$</p><p>Equation of required circle will be</p><p>\left(x + 2\right)\left(x - 2\right) +$\left(y - cot\dfrac{\theta}{2}\right)\left(y - tan\dfrac{\theta}{2}\right) = 0$</p><p>$⇒ {x}^{2} - 4 + {y}^{2} - \left(tan\dfrac{\theta}{2} + cot\dfrac{\theta}{2}\right)y + 1 = 0$</p><p>Hence, \left(\sqrt{3}, 0\right)$ satisfies the above equation.