Okay I think it is 20/3 Since A = x^2 + y^2 -8x + 6y-11 = 0, we should simplify to be (x^2-8x) + ((y^2+6x) = 11 THEN WE MAKE TRINOMIALS (add 16 and 9) x^2-8x [+16] and y^2+6x [+9] and 11+16+9 This is so we get perfect squares (x-4)^2 + (y+3)^2 = 36 Since B = 25/9 times greater than area of A (pi r^2), leave pi out and try to find the radius of B 25/9 x 36 = 25 x 4 = 100 sqroot(100) = 10 =radius of B OKAY this next part is something that YOU SHOULD GRAPH!!! draw circle A and Circle B (B>A), as they have the same center (FOR THIS CASE I WILL LABEL IT C). Then it says lines Q and R are tangent to Circle B (intersect one point = 90 degrees) and intersect point P. NOW WE HAVE A QUADRILATERAL, WITH <CQP = <CRP = 90 degrees, and <QPR = 60. TO find last angle (<QAR), we subtract from 360 (360 = 90 + 90 + 60 + x) x = 120 Okay last step is using the formular of arclength of QR and plug values in (arclength = degrees/360 x 2pi[r]) REMEMBER, the radius is from B (10) 120/360 times 2pi(10) = 1/3 times 20pi = 20/3pi Lastly since it says the arclength is kpi, we divide by pi 20/3 pi = k pi 20/3 = k IF I MADE A MISTAKE PLEASE LET ME KNOW, THIS IS HOW I DID IT AND I WOULD APPRECIATE FEEDBACK

If I really want time to pass, I would listen to lofi/classical music, or maybe just journaling random stuff that has happened

I have a duet coming up...

probably pure loyalty tbh, willing to be there in the highs, and especially the lows