Find the equation of the smaller circle tangent to both axes and passing the point (-3,6)

If the circle is tangent to both axes and passes through (-3,6), I see its centre to be in the 2nd quadrant.

If the radius is r, then its centre must be (-r,r) and its equation is
(x+r)^2 + (y-r)^2 = r^2
x^2 + 2rx + r^2 + y^2 - 2ry + r^2 = r^2
at (-3,6)
9 - 6r + r^2 + 36 - 12r = 0
r^2 - 18r + 45 = 0
(r-3)(r-15) = 0
r = 3 or r = 15

since you want the smaller circle ... r = 3
equation:
(x+3)^2 + (y-3)^2 = 9

To find the equation of the smaller circle tangent to both axes and passing through the point (-3,6), we can follow these steps:

Step 1: Determine the center of the circle.
Since the circle is tangent to both axes, the center of the circle will lie on the line y = x. Let's call the coordinates of the center (a, a).

Step 2: Find the radius of the circle.
The distance from the center of the circle to the point (-3,6) is equal to the radius of the circle.

Using the distance formula, we can calculate the distance:
r = √[(x2 - x1)^2 + (y2 - y1)^2]
= √[(-3 - a)^2 + (6 - a)^2]

Step 3: Set up the equation of the circle.
The general equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2

Therefore, the equation of the smaller circle is:
(x - a)^2 + (y - a)^2 = [√[(-3 - a)^2 + (6 - a)^2]]^2

This equation represents the circle that is tangent to both axes and passes through the point (-3,6), with its center at (a, a).

To find the equation of the smaller circle that is tangent to both axes and passes through the point (-3,6), we can follow these steps:

Step 1: Find the equation of the larger circle
Since the smaller circle is tangent to both axes, it will share a common center with the larger circle. As the smaller circle passes through (-3,6), the center of the larger circle will also be the midpoint of the line segment joining the point (-3,6) and the origin (0,0), which is (0,3) (the point equidistant from the x-axis and the y-axis).

The distance between the center of the larger circle and any point on the circle will be the radius. We can choose the distance between the center and the point (-3,6) as the radius. Using the distance formula:

r = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((0 - (-3))^2 + (3 - 6)^2)
= sqrt(3^2 + (-3)^2)
= sqrt(9 + 9)
= sqrt(18)
= 3√2

Therefore, the equation of the larger circle is (x - 0)^2 + (y - 3)^2 = (3√2)^2.

Simplifying, we get:
x^2 + (y - 3)^2 = 18

Step 2: Find the equation of the smaller circle
Since the smaller circle is internally tangent to the larger circle, the radius of the smaller circle will be the difference between the radius of the smaller circle and the radius of the larger circle. Let's denote the radius of the smaller circle as r_s.

So, the radius of the smaller circle is r_s = 3√2 - 3√2 = 0.

Since the radius is 0, the equation of the smaller circle will be x^2 + y^2 = 0.

Therefore, the equation of the smaller circle that is tangent to both axes and passes through the point (-3,6) is x^2 + y^2 = 0.