Posted by **Lance** on Monday, January 7, 2013 at 4:57pm.

What are the equations of the lines through (-5,-3) and passing at distance 2sqrt5 from (5,7)

Please I really need your help. I do not know how to do this problem.

- Math,Geometry -
**Reiny**, Monday, January 7, 2013 at 6:36pm
Make a sketch, let the point of contact be P(x,y)

(there will be two of them, let the math take care of it)

label A(-5,-3) and B(5,7)

point P must be on a circle with centre B(7,5) and radius of BP = 2√5

the equation of that circle is

(x-7)^2 + (y-5)^2 = 20

which when expanded gets us

**x^2 + y^2 = 10x+14y-54**

also the slope of AP must be the negative reciprocal of the slope of BP

slope AP = (y+3)/(x+5)

slope BP = (y-7)/(x-5)

then (x-5)/-(y-7) = (x+5)/(y+3)

x^2 - 25 = -y^2 + 7y + 21 - 3y

**x^2 + y^2 = 4y + 46**

So 10x + 14y - 54 = 4y+46

10x = -10y + 100

**x = 10- y**

Then in **x^2 + y^2 = 4y + 46**

(10-y)^2 + y^2 = 4y + 46

100 - 20y + y^2 + y^2 = 4y + 46

2y^2 - 24y + 54 = 0

y^2 - 12y + 27 = 0

(y-3)(y-9)= 0

y = 3 or y = 9

if y = 3, then x = 10-3 = 7 ...... P is (7,3)

if y = 9, then x = 10-9 = 1 ...... P is (1,9)

almost done ......

for P as (7,3)

slope of AP = (3+3)/(7+5) = 6/12 = 1/2

equation of AP:

y+3 = (1/2)(x+5)

2y + 6 = x+5

x - 2y = 1 is one of the line equations

for P as (1,9)

.......

I will let you have the pleasure of finishing it.

(check my work, but it came out so nice to have an error in it)

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