Question

How do I find the points of the line x+7y-52=0 on a distance of 5 from point (6,3) ?

Can you explain in detail how I get there ?

Answer 1

`(x,y) in {(17,5),(-4,8)}`

Explanation:

If `x+7y-52=0`
then
`color(white)("XXX")x=52-7y`

A circle with center `(6,3)` and radius `5` has an equation
#color(white)("XXX")(x-6)^2+(y-3)^2=5^2

Substituting `52-7y` for `x` in the equation o0f the circle gives
`color(white)("XXX")((52-7y)-6)^2+(y-3)^2=25`

`color(white)("XXX")(46-7y)^2+(y-3)^2=25`

`color(white)("XXX")(1996-644y+49y^2)+(y^2-6y+9)=25`

`color(white)("XXX")50y^2-650y+2005=25`

`color(white)("XXX")2y^2-26y+81=1`

`color(white)("XXX")2y^2-26y+80=0`

`color(white)("XXX")y^2-13y+40=0`

which factors as
`color(white)("XXX")(y-5)(y-8)=0`

that is
`color(white)("XXX")`
`{:(y=5," or ",y=8),(rarr x=52-7xx(5),,rarrx=52-7xx(8)),(rarrx=17,,rarrx=-4):}`