Test:1 QNo:5

i dont exactly know how to solve it, but i know that negative values of x anf y will maximise the expression. for example, give x=-1 and y=-2, then, expression will get a value of 32. giving x=-3 and y=0, the exprssion will gain a value of 36.

The given condn is a circle with centre (-3,-2) and radius 2.The expression given is thus the maximum distance of a point on the circle from (3,2).The answer is 2+√13.

Let us take an arbitrary pt. on the circle
x+3=2cosA & y+2=2sinA or, x=2cosA-3 & y=2sinA-2
Reqd, max of F(x)=(x-3)²+(y-2)²
or, F(x)=(2cosA-6)²+(2sinA-4)²
or, F(x)=56-8(3cosA+2sinA)
min. of 3cosA+2sinA=-√13
so, max F(x)=56+8√13

actually using graph is easier

now eqn of AB : 3y = 2x

find pt of intersection of line AB and given circle , u get 2 set of (x,y) , take the one with more negative values of x,y i.e pt A , and find square of distance frm 3,2