Posted by **geometry** on Monday, January 24, 2011 at 2:18am.

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

- gggg -
**Reiny**, Monday, January 24, 2011 at 8:45am
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.

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