Let the lengths of intercepts on x -axis and y -axis made by the circle x 2 + y 2 + a x + 2 a y + c = 0 , a < 0 be 2 2 and 2 5 , respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2 y = 0 , is equal to :

Step-by-step solution

x 2 + y 2 + a x + 2 a y + c = 0 2 g 2 - c = 2 a 2 4 - c = 2 2 ⇒   a 2 4 - c = 2   . . . 1 2 f 2 - c = 2 a 2 - c = 2 5 ⇒ a 2 - c = 5   . . . 2 1   &   2 3 a 2 4 = 3 ⇒ a = - 2   a < 0 ∴    c = - 1 Circle ⇒ x 2 + y 2 - 2 x - 4 y - 1 = 0 ⇒ x - 1 2 + y - 2 2 = 6 Given x + 2 y = 0 ⇒ m = - 1 2 m tangent = 2 Equation of tangent ⇒ y - 2 = 2 x - 1 ± 6 1 + 4 ⇒ 2 x - y ± 30 = 0 Perpendicular distance from 0 , 0 = &#1