The equation of a circle passing through point (8,1) and tangent to two coordinate axes

If it is tangent to both coordinate axes, the distance from the center of the circle to the two coordinate axes is equal. If the radius is r, then a = B or a = - Br = | a | then (x-a) ^ 2 + (Y-A) ^ 2 = a ^ 2 or (x-a) ^ 2 + (y + a) ^ 2 = a ^ 2. Substitute (8,1) into (x-a) ^ 2 + (Y-A) ^ 2 = a ^ 2 (8-A) ^ 2 + (1-A) ^ 2 = a ^ 2A ^ 2-16a + 64 + A ^ 2-2a + 1 = a ^ 2-18a + 65 = 0

The equation of the circle whose two coordinate axes are tangent and pass through the point (2,1),
The equation of a circle whose two axes are tangent and pass through point (2,1), why?

Obviously, the circle passing through point (2,1) is in the first quadrant
Let the equation of the circle be (x-a) & sup2; + (Y-A) & sup2; = A & sup2;, a > 0
Substituting point (2,1), we get (2-A) & sup2; + (1-A) & sup2; = A & sup2;
That is a & sup2; - 6A + 5 = 0
That is, (A-1) (a-5) = 0
We get a = 1 or a = 5
The circle is (x-1) & sup2; + (Y-1) & sup2; = 1 or (X-5) & sup2; + (Y-5) & sup2; = 25

The equation of a circle passing through point (2,1) and tangent to both axes is ()
A. (x-1) 2 + (Y-1) 2 = 1b. (x-1) 2 + (Y-1) 2 = 1 or (X-5) 2 + (Y-5) 2 = 5C. (x-1) 2 + (Y-1) 2 = 1 or (X-5) 2 + (Y-5) 2 = 25d. (X-5) 2 + (Y-5) 2 = 5

Let (x-a) 2 + (Y-A) 2 = A2 (2,1) be substituted into the equation of circle, then: (2-A) 2 + (1-A) 2 = A2 (a-5) × (A-1) = 0A = 5 or 1 (X-5) 2 + (Y-5) 2 = 25 (x-1) 2 + (Y-1) 2 = 1

The equation of a circle passing through point (8,1) and tangent to two coordinate axes