The line x + 2y - 10 = 0 is a tangent to a circle whose centre is at point (2,-1). Find the tangent point and the second tangent point of a line with the same

Question

The line x + 2y - 10 = 0 is a tangent to a circle whose centre is at point (2,-1). Find the tangent point and the second tangent point of a line with the same slope as the given line.

Answer 1

tangent points are (4,3) and (0,-5)

x+2y-10=0 rewrite as y=(-1/2)x+5. That line has a y-intercept at y=5, and a slope of -1/2

the radius of the circle with center (2,-1) is the same as the perpendicular from the line to the point.

That perpendicular line has slope of 2, the negative inverse of the original line

Use the point slope formula to find the equation of the perpendicular line

y+1 = 2(x-2)

y= 2x-4-1= 2x-5

Set y=y or 2x-5 = (-1/2)x+5 to determine the point of intersection of the two lines

2x+(1/2)x=10

(5/2)x=10

x=4

plug 4 into the equation for either line to get the y value

y=2x-5=2(4)-5=8-5=3

(4,3) is the intersection point of the perpendicular and the original line

Find the distance from the center of the circle, (2,-1) to (4,3)

d = square root of (4-2)² + (3+1)² = square root of 20 = radius

The other line tangent to the circle, parallel to the original line, has the same slope = 2

The equation of the circle is (x-2)²+(y+1)² = r² = 20

It touches the parallel line where the perpendicular line or radius intersects the parallel line

the perpendicular line is y=2x-5 and the parallel line is y=(-1/2)x-5

set those equal and solve for x 2x-5 = (-1/2)x-5 x=0, y=-5, the tangency point is (0,-5)