Question

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

This problem is confusing? How do I solve this? Please I need your Help

Answer 1

all points 2√5 from (5,7) form a circle:

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

Now you want a tangent to the circle that passes through (-5,-3). Must be a tangent, because any other line through the circle will come closer than 2√5 to the center.

So, since lines through (-5,-3) with slope m are

(y+3) = m(x+5), we need

(x-5)^2 + (m(x+5)-3-7)^2 = 20
(x-5)^2 + (mx+(5m-10))^2 = 20
(m^2+1)x^2 + 10m(m-3)x + (5m-10)^2 - 20 = 0

Now for the line to be tangent, the above equation must have a single root. That is, the discriminant must be zero:

100m^2(m-3)^2 - 4(m^2+1)((5m-10)^2-20) = 0
m = 1/2, 2

so, the two lines are

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