Given:
The centre of the circle is given as (h, k) = (2,2)
We know that the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).
r = \( \sqrt{(2-4)^2 + (2-5)^2}\)
= \( \sqrt{(-2)^2 + (-3)^2}\)
= \( \sqrt{4+9}\)
= \( \sqrt{13}\)
The equation of the circle is given as
(x– h)2+ (y – k)2 = r2
(x –h)2 + (y – k)2 = \( ( \sqrt{13})^2\)
(x –2)2 + (y – 2)2 = \( ( \sqrt{13})^2\)
x2 – 4x + 4 + y2 – 4y + 4 = 13
x2 + y2 – 4x – 4y = 5
The equation of the required circle is x2 + y2 – 4x – 4y = 5
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