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)
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.
1 answer
1 answer