Question

What are the two points of intersection of the line 3y+2=4x and the circle with the radius 5 and center (5,6)

Answer 1

first, fidn equation

for a circle with center (h,k) and radius, r the equation is

`(x-h)^2+(y-k)^2=r^2`
center (5,6) and radius 5

`(x-5)^2+(y-6)^2=5^2`
expand

`x^2+y^2-10x-12y+36=0`

now

3y+2=4x
divide both sides by 4
3/4y+1/2=x

sub that for x


`(3/4y+1/2)^2+y^2-10(3/4y+1/2)-12y+36=0`

`(9/16y^2-3/4y+1/4+y^2-30/4y+5-12y+36=0`
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)