Question

Find the equation of a circle with center (2, 2) and passes through the point (4, 5).

Answer 1

The center of the circle is given as $(h, k) = (2, 2).$
Since the circle passes through points $(4, 5),$ the radius ($r$) of the circle is the distance between the points $(2, 2)$ and $(4, 5).$

$r = √[(2-4)^2 + (2-5)^2]$$$

$= √[(-2)^2 \+ (-3)^2]$

$= √[4+9]$

$= √13$

Thus, the equation of the circle is

$(x– h)^2+ (y – k)^2 = r^2$

$(x –h)^2 \+ (y – k)^2 = (√13)^2$

$(x –2)^2 \+ (y – 2)^2 = (√13)^2$

$x^2 – 4x + 4 + y^2 – 4y + 4 = 13$

$x^2 \+ y^2 – 4x – 4y = 5$

∴ The equation of the required circle is $x^2 \+ y^2 – 4x – 4y = 5$ (Ans.)