Find the equation of the locus of the mid-points of chords of the circle 2x²+ 2y² - 6x + 4y + 3 = 0 subtending an angle pi/2 at its centre

let  the end points of chord be (x1,y1) and (x2,y2).
mid point: (x,y)= ((x1+x2)/2),(y1+y2)/2)

centre= (3/2,-1) and radius is 1/2.
Since, the angle subended at the centre is of 90.
So, distance between the end-points of chord will be sqrt(2). (check diagramatically).
so, (x1-x2)^2 + (y1-y2)^2 = 1/2.
put the value of x1= 2x-x2 and similarly for y1.
so, we get (x-x2)^2 + (y-y2)^2 = 1/8.

now, join centre and mid point, and see carefully that angles in each half are two angles 45 and 1 angle is of 90.
apply, pythogoras theorem in half section.
((x-3)^2 + (y+2)^2)  + ((x-x2)^2 + (y-y2)^2) =1/4.

solve both equations:
u'll get x^2 -6x +y^2 +4x +13 =1/8.

If still you have any doubt do tell me...

well u can solve it also in following method

see ,,,,let the mid point of chord be (h,k)

so equation of chord when mid point is known can be written as

T=S1............................(1)

now u have the equation of circle and the eqn(1),,,,so with the help of homogenisation principle

homogenise thge eqn of curve  using (1)

and then u will get the 2nd degree eqn,, as chord is subtending angle pi/2

so   homogenised eqn  will represent the line joining the centre and the points where chord cut the circle

for angle b/w those lines to be pi/2

coeff of x^2+coeff of y^2=0    ....(of homogenised eqn)

u will get the locus in h and k,,,at last replace them with x and y

hope i helped u