If the center of the circle is on the x-axis and passes through the point (3, root 3) (0,0), find the equation of similar circle

If the center of the circle is on the x-axis and passes through the point (3, root 3) (0,0), find the equation of similar circle

Let the center of the circle be (x, 0)
(x-3) ^ 2 + (0-radical 3) ^ 2 = (x-0) ^ 2 + (0-0) ^ 2
So, x = 2
r^2=(2-0)^2+(0-0)^2=4
So, the equation of the circle is, (X-2) ^ 2 + y ^ 2 = 4

The equation of the circle with the length of 5 root sign 2 obtained by taking the point (1, - 3) as the center and the tangent line x + 7y-5 = 0

Let the equation of circle be: (x-1) ^ 2 + (y + 3) ^ 2 = R ^ 2 (r > 0 is radius)
Because the distance d from the center of the circle to the straight line, the half chord length and the radius can form a right triangle (drawing), it can be obtained from the Pythagorean theorem:
d^2+(l/2)^2=r^2
Because d = | 1-21-5 | / √ (1 + 7 ^ 2) = 5 √ 2 / 2
And L / 2 = 5 √ 2 / 2
So R ^ 2 = 25
So the equation of circle is: (x-1) ^ 2 + (y + 3) ^ 2 = 25