CircleMedium

Question

An equilateral triangle OAB is inscribed in the parabola $y^{2} = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is

Options

A.$4(3 - \sqrt{3})$
B.$2(8 - 3\sqrt{3})$
C.$4(6 + \sqrt{3})$
D.$2(3 + \sqrt{3})$

Solution

$${M_{OA} = \frac{2t - 0}{t^{2} - 0} = \frac{2}{f} }{\frac{2}{t} = tan30^{\circ} }{t = 2\sqrt{3} }$$Req. Circle : $(x - 12)^{2} + y^{2} = (4\sqrt{3})^{2}$

Least distance $= |CP - R|$

$$= \mid 2 - 4\sqrt{3} = 4(3 - \sqrt{3})$$

Create a free account to view solution

View Solution Free
Topic: Circle·Practice all Circle questions

More Circle Questions

Equation (2 + λ)x2 - 2λxy + (λ - λ)y2 - 4x - 2 = 0 represents a hyperbola if -...Circles x2 + y2 − 2x − 4y = 0 and x2 + y2 − 8y − 4 = 0...In the argand plane the inequality |(√3 + i)z - (√2 - i)|2 + |(√ + i)z + (√3 - i) |2...The equation of the chord of the circle x2 + y2 − 6x + 8y = 0 which is bisected at the point (5, −3) is-...If latus rectum of an ellipse = 1 {0 < b < 4}, subtend angle 2θ at farthest vertex such that cosec θ = &...