Answer:
3x-2y-1=0
or, 3x-2y=1
or, x=(1+2y)/3...(i)
4x+y=27... ...(ii)
solving (i) and (ii)
(4+8y)+3y=27x3
4+11y= 81
11y=77
y=7
now,
x=(1+2x7)/3
x=5
hence, the passing point is (5,7)
now, for the radius of circle
radius of circle= distance betn (5,7) and (2,3)
r= √(5-2)²+(7-3)²
r=5units
finally,
the eqn of circle
(x-h)²+(y-k)²=r² (where (h,k)=(2,3) )
(x-2)²+ (y-3)²=5²
x²-4x+4+y²-6y+9=25
x²+y²-4x-6y=12....this is the required answer.
