first, fidn equation
for a circle with center (h,k) and radius, r the equation is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
center (5,6) and radius 5
[tex](x-5)^2+(y-6)^2=5^2[/tex]
expand
[tex]x^2+y^2-10x-12y+36=0[/tex]
now
3y+2=4x
divide both sides by 4
3/4y+1/2=x
sub that for x
[tex](3/4y+1/2)^2+y^2-10(3/4y+1/2)-12y+36=0[/tex]
[tex](9/16y^2-3/4y+1/4+y^2-30/4y+5-12y+36=0[/tex]
after simplification and factorization we get
(25/16)(y-10)(y-2)=0
set to zero
y-10=0
y=10
y-2=0
y=2
sub and solve
3/4y+1/2=x
3/4(10)+1/2=x
30/4+2/4=x
32/4=x
8=x
(8,10) is one oint
3/4y+1/2=x
3/4(2)+1/2=x
6/4+2/4=x
8/4=x
2=x
(2,2)is another point
the points are
(8,10) and (2,2)
