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
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
To solve this problem, you can use the distance formula to find the equation of a line passing at a certain distance from a given point.
Step 1: Find the distance between the two given points.
- The given points are (-5, -3) and (5, 7).
- The distance formula is given by: d = √[(x2 - x1)^2 + (y2 - y1)^2].
- Plugging in the values, we get: d = √[(5 - (-5))^2 + (7 - (-3))^2].
- Simplifying, we get: d = √[(10)^2 + (10)^2] = √[100 + 100] = √200 = 10√2.
Step 2: Determine the equations of the lines passing through (-5, -3) and at distance 2√5 from (5, 7).
- Use the point-slope form of the equation of a line: y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope of the line.
- The line passing through (-5, -3) can be represented as: y - (-3) = m(x - (-5)), which simplifies to y + 3 = m(x + 5).
- The line passing at a distance of 2√5 from (5, 7) can be represented as: (y - 7)^2 + (x - 5)^2 = (2√5)^2.
- Expanding and simplifying, we get: y^2 - 14y + 49 + x^2 - 10x + 25 = 20.
- Rearranging, we get: x^2 - 10x + y^2 - 14y + 4 = 0.
Therefore, the equations of the lines passing through (-5, -3) and at a distance of 2√5 from (5, 7) are y + 3 = m(x + 5) and x^2 - 10x + y^2 - 14y + 4 = 0.