COORDINATE GEOMETRY -----------------------------------  WANT FREE COACHING IN MATHS IN BANGALORE ? CALL 9480193388 OR EMAIL freemaths10212@gmail.com For details see post on 26-Feb-2015 at 6:00 am. ------------------------------------- If (x, y) is inside the circle (x - 3)^2 + (y - 2)^2 = 4 then x-6<3y

COMPARING WITH STD.EQN.

(X-H)^2+(Y-K)^2=R^2,WE HAVE

CENTRE OF CIRCLE = O = (H,K) =(3,2).....RADIUS = R =2

FOR A POINT P = (X,Y) TO BE INSIDE THE CIRCLE,ITS DISTANCE FROM CENTRE SHOULD BE LESS THAN THE RADIUS

TPT...IF P IS INSIDE THE CIRCLE THEN

X-6<3Y

TPT....3Y>X-6...OR.....Y>X/3 -2...OR......Y-X/3 +2>0.

TEST C ON THIS LINE CONDITION....2-3/3+2=4...THAT IS CENTRE SATISFIES THIS CONDITION.LET US FIND SHORTEST DISTANCE OF CENTRE FROM THE LINE Y-X/3 +2=0

S.D.=(2-1+2)/(1+1/9)^0.5=9/10^0.5=2.846>R OF 2

THAT IS P SHOULD BE OVER THE LINE(THAT IS THE LINE SHOULD BE BETWEEN ORIGIN AND POINT) Y-X/3 +2