How do you find the center of the circle with a radius of 5, tangent to the line y = x and the x-axis?

Question

Assume the center is in the first quadrant.

Answer 1

Let the center of the circle be (x0,y0). Then, as it is of radius 5 and tangent to the x-axis, y0 = 5.

Consider the triangle (5,5), (x0, 5), (x0/2+5/2,x0/2+5/2). This will be isosceles, and a right triangle as the tangent is perpendicular to the radius. Thus, the hypotenuse is equal to x0 - 5 and each side has length 5. Thus, the hypotenuse has length 5√2, so x0 - 5 = 5√2, and x0 = 5 + 5√2.

The center is (5 + 5√2, 5)

We will also, of course, have (-5 - 5√2, -5)

Answer 2

The prerpendicular distance from the center to the x axis and the line y=x must be 5 (the radius)

So the center must have y coordinate 5.

Suppose it is (a,5).

Lines perpendicular to y=x have slope -1, so form y = -x+b

Such a line passing through (a,5) gives 5 = -a + b so b = a+5

So y = -x + a + 5. this intersects y=x when x = (a+5)/2, so at the point ( (a+5)/2, (a+5)/2)

Basically the distance between this point and (a,5) has to be 5. You can use the distance formula to solve this for a, giving a = 12.1

So the circle center is (12.1, 5). You can plot this circle of Desmos [ (x-12.1)² + (y-5)² = 5² and check that y=x and x axis are tangents]