what are the equation of a line through (8,-3) and passing distance square root of 10 from (3,2)

Answer 1

To be a "distance of √10 units away from (3,2) means it has to be tangent to the circle
(x-3)^2 + (y-2)^2 = 10
or
x^2 + y^2 -6x - 4y = -3

let the point of contact be (a,b)
then
a^2 + b^2 - 6a - 4b = -3 (#1)

by slopes...
slope of radius to (a,b) = (b-2)/(a-3)
slope from (a,b) to (8,-3) = (b+3)/(a-8)
but these two lines are perpendicular, so ...
(b+3)/(a-8) = - (a-3)/(b-2)
which simplifies to
a^2 + b^2 - 11a + b = -18 ,(#2)

subtract #1 - #2
5a - 5b = 15
a - b = 3
a = b+3 (#3)

sub #3 into #2 to get it simplified to
b^2 - 2b - 3 = 0
(b-3)(b+1) = 0
b = 3 or b = -1
in #3
a= 6 or a = 2

so (a,b) is either (6,3) or (2,-1)

Now it is easy to find the equation of the two tangents.
Let me know what you got.

Answer 2

3x+y-21=0 and x+3y+1