ParabolaMedium
Question
From the focus of the parabola y2 = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -
Options
A.(x - 2)2 + y2 = 3
B.(x - 2)2 + y2 = 9
C.(x + 2)2 + y2 = 9
D.x2 + y2 - 4x = 0
Solution

Focus of parabola y2 = 8x is (2, 0). Equation of circle with centre (2, 0) is
(x - 2)2 + y2 = r2
AB is common chord
Q is mid point i.e. (1, 0)
AQ2 = y2 where y2 = 8 × 1 = 8
∴ r2 = AQ2 + QS2 = 8 + 1 = 9
so circle is (x - 2)2 + y2 = 9
Create a free account to view solution
View Solution FreeMore Parabola Questions
Let the point P be the vertex of the parabola $y = x^{2} - 6x + 12$. If a line passing through the point $P$ intersects ...F1 and F2 are focus of hyperbola 16x2 - 9y2 - 32x = 128.S = 0 is equation of parabola whose vertex is F1 and focus is F2...If the tangent at the point P (x1, y1) to the parabola y2 = 4ax meets the parabola y2 = 4a (x + b) at Q & R, then the mi...Equation of a common tangent to the circle, $x^{2} + y^{2} = 50$ and the parabola, $y^{2} = 40x$ can be :...Consider a branch of the hyperbola x2 - 2y2 - 2 √2x - 4√2y - 6 = 0 with vertex at the point A. Let B be one ...