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.
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)