# Equation of circle 2

Find the equation of a circle which touches the axis of y at a distance 4 from the origin and cuts off an intercept of length 6 on the axis x.

Result

f = (Correct answer is: )

#### Solution:

$A=(0,4) \ \\ B=(x_{0}+6/2,0) \ \\ C=(x_{0}-6/2,0) \ \\ \ \\ (x-x_{0})^2 + (y-y_{0})^2=r^2 \ \\ \ \\ (0-x_{0})^2 + (4-y_{0})^2=r^2 \ \\ (x_{0}+3-x_{0})^2 + (0-y_{0})^2=r^2 \ \\ (x_{0}-3-x_{0})^2 + (0-y_{0})^2=r^2 \ \\ \ \\ x_{0}^2 + (4-y_{0})^2=r^2 \ \\ 9 + y_{0}^2=r^2 \ \\ 9 + y_{0}^2=r^2 \ \\ \ \\ r=5 \ \\ x_{0}=5 \ \\ y_{0}=4 \ \\ \ \\ f=(x-5)^2+(y-4)=5^2$

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