Someone help me with this ASAP

here's the solution,

let the segment be AB, and PQ is perpendicular bisector of AB at Q , and Let P be the point on perpendicular bisector, and join P with A and B,

and since PQ bisects AB into AQ and BQ, they are equal,

that is , AQ = BQ

now, two triangles are formed.

In the given triangles, PQA and PQB

( common side )

( perpendicular forms 90° )

( they are equal, proved above )

here, Triangle PQA and PQB are congruent by SAS ( side angle side ) criterion.

So, AP = BP ( by C.P.C.T )

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.