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

User Olliepower
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1 Answer

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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).

User Ateeb
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