The intercepts of the circle (x-1) ^2 + (y-2) ^2 = 10 are?

y-intercept: Let x = 0 and solve for y:

(x-1) ^2 + (y-2) ^2 = 10 = > (-1) ^2 + y^2 - 4y + 4) = 10

=> 1 + y^2 - 4y + 4 = 10, or y^2 - 4y - 5 = 0

The solutions of this quadratic are y = 5 and y = - 1.

Thus, the y-intercepts are (0, 5) and (0, - 1).

Now find the x-intercepts: Let y = 0 and solve the resulting equation for x:

(x-1) ^2 = 10 - (-2) ^2, or (x-1) ^2 = 10 - 4 = 6.

Taking the sqrt of both sides, x - 1 = plus or minus sqrt (6), or:

x = 1 + √6 and x = 1 - √6, so that the x-intercepts

are (1+√6, 0) and (1-√6, 0).