# 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:

$A=(0,4) \ \\ B=(x_{0}+6/2,0) \ \\ C=(x_{0}-6/2,0) \ \\ \ \\ (x-x_{0})^2 + (y-y_{0})^2=r^2 \ \\ \ \\ (0-x_{0})^2 + (4-y_{0})^2=r^2 \ \\ (x_{0}+3-x_{0})^2 + (0-y_{0})^2=r^2 \ \\ (x_{0}-3-x_{0})^2 + (0-y_{0})^2=r^2 \ \\ \ \\ x_{0}^2 + (4-y_{0})^2=r^2 \ \\ 9 + y_{0}^2=r^2 \ \\ 9 + y_{0}^2=r^2 \ \\ \ \\ r=5 \ \\ x_{0}=5 \ \\ y_{0}=4 \ \\ \ \\ f=(x-5)^2+(y-4)=5^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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