The radius of the circle passes through the foci of a conic x^2 16 + y^2 9 =1 and has its centre at (0,3) , then the diameter of the circle is ---

Step-by-step solution

The given equation of the ellipse is x^2 16 + y^2 9 = 1 . Here a^2 = 16 and b^2 = 9 . The eccentricity e is given by e = 1 - b^2 a^2 = 1 - 9 16 = 7 16 = 7 4 . The foci of the ellipse are ( ae, 0) . Substituting the values, ae = 4 7 4 = 7 . Thus, the foci are F₁ = ( 7 , 0) and F₂ = (- 7 , 0) . The circle passes through these foci and has its center at C = (0, 3) . The radius r of the circle is the distance between the center (0, 3) and either focus ( 7 , 0) . r = ( 7 - 0)^2 + (0 - 3)^2 = 7 + 9 = 16 = 4 . The diamete