find the equation of the circle with centre (2,3) and passing through the intersection of the lines 3x-2y-1=0 and 4x+y-27=0

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=12this is the required answer.

use formula for the circle a=3.14

step-by-step explanation:

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.