IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

aditi_g

find the locus of the centres of the cirscles passing through the intersection of the circles x^2 + y^ 2 = 1 and x^2 + y^2 - 2x + y=0

i lknow the answer but i need the method to solve such kinds of questions in circles

joyfrancis

Find the point of intersection of the two circles first.

x²+y²=1

x²+y²=2x-y

2x-1=y....(1)

put value of y in first equation

x²+(2x-1)²=1

=>x(5x-4)=0

=> x=0 and x=4/5

the corresponding values of y come out to be -1 and 3/5

Now,

we have to find the locus of the center of the circles which pass through (0,-1) & (4/5,3/5)

Let center be (h,k)

.: (h-0)²+(k+1)²=(h-(4/5))²+(k-(3/5))²

=> 2h=k

So, locus is 2x=y.....(ans)

rhd92781

any circle passing throught the intersection of 2 circles S1 & S2 =0 is S1+

S2=0

so x²+y²-2x+y+

(x²+y²)=0

its centre is (1/

+1, -1/2(

+1))

so h=1/(

+1) and k=-1/2(

+1)

so

+1 = 1/h = -1/2k

or h=-2k or h+2k=0

so locus is x+2y=0

aditi_g

hey joy thr was some error in ur answer.....neways thanx for taking out ur time to solve it