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posted by geometry on .
what are the equation of a line through (8,3) and passing distance square root of 10 from (3,2)

To be a "distance of √10 units away from (3,2) means it has to be tangent to the circle
(x3)^2 + (y2)^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) = (b2)/(a3)
slope from (a,b) to (8,3) = (b+3)/(a8)
but these two lines are perpendicular, so ...
(b+3)/(a8) =  (a3)/(b2)
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
(b3)(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. 
3x+y21=0 and x+3y+1