What is the equation of the circle with the center O(2,-1) that is tangent to the straight line r: y=x+2?

Question

Please include your solution. Thank you.

Answer 1

Razel:

If y=x+2 is tangent to the circle with center O(2,-1), the radius of the circle is equal to the distance between the center of the circle and the line y=x+2 or y-x-2=0.

The distance between a point and a line is |ax₀+by₀+c| / √(a²+b²)

where the line has equation ax+by+c =0 and the point is at (x₀,y₀)

We have y-x-2=0 and O(2,-1). So a = -1, b = 1, c = -2, x₀ = 2, y₀ = -1. Plugging in those values yields

distance =5/√2.

The circle has equation (x-x₀)²+(y-y₀)² = d² Plug in x₀ y₀ and d.

Answer 2

Radius r of circle is distance from center O(2,-1) to the tangent line y = x + 2. To get this distance we change equation of tangent line to normal form: x - y + 2 = 0, then deviding this equation by ((1)² + (-1)²)^1/2 = sqrt(2).

So normal equation is (x - y)/sqrt(2) + sqrt(2) = 0. And r is equel absolute value of result of substitution of coordinates of O(2,-1) in normal equation: r = abs((2 + 1)/sqrt(2) + sqrt(2)) = 5/2sqrt(2).

Then equation of circle: (x - 2)² + (y + 1)² = 25/2