Question

Ferris wheel cars are at a position of A (9,33) and B (27,-15). One of its axle is

represented by 3y+2x-7= 0 where it passes through the center of the wheel.

Draw the suitable graph of the said situation.

Answer 1

$(x-a)^2 + (y-b)^2 =r^2$

$(9-a)^2 + (33-b)^2 =r^2$

$(27-a)^2 + (-15-b)^2 =r^2$

Eliminating $r^2$

$((9-a)^2 + (33-b)^2) -((27-a)^2 + (-15-b)^2)=0$

$(81-18a+a^2 + 1089-66b+b^2)-(729-54a+a^2+225+30b+b^2)=0$$81+1089-729-225-18a+54a-66b-30b +a^2-a^2+b^2-b^2=0$$216+36a-96b=0$

dividing through by 12 yields

$3a-8b+18=0$

Axle equation:

$3y+2x-7=0.$

it passes at $(a,b)$

hence

$(3a-8b+18=0).2$

$(2a+3b-7=0).3$

$6a+9b-21=0$

$-$

$6a-16b+36=0$

$25b-57=0$

$b=57/25$

$2a+3*(57/25)-7=0$

$2a-4/25=0$

$2a=4/25$

$a=2/25$

Thus the center of the wheel is at $(2/25, 57/25)$

Equation of the wheel becomes$(x-2/25)^2+(y-57/25)^2=r^2$

now we must get radius r

one point on the circumference is $A(9,33)$ ,center is $C(2/25, 57/25)$

thus the radius is the magnitude of the distance between A and C

$r^2= ((9-2/25)^2+(33-57/25)^2)=1023.2848$

$r=32 units$

hence equation of the circle becomes

$(x-2/25)^2+(y-57/25)^2=32^2$

Sketching this gives: