coordinate geometry(circle)

The circle touches both axis and passes through the point (-2, 1).
Since (-2, 1) is in the 2nd quadrant, thus the circle lies on the 2nd quadrant.


Let (-a, a) be the centre of circle, where a > 0.
Then, radius of the circle = a

Equation of the circle:
(x + a)² + (y - a)² = a²
x² + 2ax + a² + y² - 2ay + a² = a²
x² + y² + 2ax - 2ay + a² = 0

(-2, 1) lies on the circle:
(-2)² + (1)² + 2a(-2) - 2a(1) + a² = 0
4 + 1 - 4a - 2a + a² = 0
a² - 6a + 5 = 0
(a - 1)(a - 5) = 0
a = 1 or a = 5

When a = 1, the equation of the circle :
x² + y² + 2(1)x - 2(1)y + (1)² = 0
x² + y² + 2x - 2y + 1 = 0

When a = 5, the equation of the circle :
x² + y² + 2(5)x - 2(5)y + (5)² = 0
x² + y² + 10x - 10y + 25 = 0