If is the perpendicular distance of a point P on the circle x^2+y^2+2 x+2 y-3=0 , from the line 2 x+y+13=0 , then maximum possible value of is

Step-by-step solution

Given equation of the circle is x^2+y^2+2 x+2 y-3=0 Which can be written as: (x+1)^2+(y+1)^2=5 It is a circle with centre (-1,-1) and radius 5 Given line is: 2 x+y+13=0 To find the required distance, we find the equation of a line perpendicular to the given line, and passing through the centre of the given circle. Equation of this line is: (y+1)= 1 2 (x+1) i.e., x=2 y+1 Now, we find the points where line x=2 y+1 intersects the circle x^2+y^2+2 x+2 y-3=0 array ll & (2 y+1)^2+y^2+2(2 y+1)+2 y-3=0 & 4 y^2+4 y+1+y^2+4