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).

d. vertex

step-by-step explanation: