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

First, fidn equation

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

center (5,6) and radius 5

expand


now

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

sub that for x



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)

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12