Let the circle with radius R having its centre at O be x^2+y^2=R^2. Now OP=RS=R. Since areas A and B are equal, we've Pi*R^2/4 -(Commonn Area)=R*OS -(Common Area). In other words, OS = RT = Pi*R/4 and OR^2=(PiR/4)^2+R^2 Hence, OR =(R/4)V(18+Pi^2). Now let T be (a,b). We've R as (Pi*R/4,R) Since T is a point on the circle we can say: a^2 + b^2 = R^2 & ((pR/4)-a)^2+(R-b)^2=p^2*R^2/16 On solving we can obtain a & b and since S is (PiR/4,0) we can show that ST =R(16-Pi^2)/(4V(16+Pi^2). Hence ST/OR =(16-Pi^2)/(16+Pi^2) Ajit
Let the circle with radius R having its centre at O be x^2+y^2=R^2. Now OP=RS=R. Since areas A and B are equal, we've Pi*R^2/4 -(Commonn Area)=R*OS -(Common Area). In other words, OS = RT = Pi*R/4 and OR^2=(PiR/4)^2+R^2 Hence, OR =(R/4)V(18+Pi^2). Now let T be (a,b). We've R as (Pi*R/4,R) Since T is a point on the circle we can say: a^2 + b^2 = R^2 & ((pR/4)-a)^2+(R-b)^2=p^2*R^2/16 On solving we can obtain a & b and since S is (PiR/4,0) we can show that ST =R(16-Pi^2)/(4V(16+Pi^2). Hence ST/OR =(16-Pi^2)/(16+Pi^2)
ReplyDeleteAjit
Your approach of analytic geometry is interesting and very good!
ReplyDeleteThanks for the solution.