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: ) OK

Solution:

A=(0,4) B=(x0+6/2,0) C=(x06/2,0)  (xx0)2+(yy0)2=r2  (0x0)2+(4y0)2=r2 (x0+3x0)2+(0y0)2=r2 (x03x0)2+(0y0)2=r2  x02+(4y0)2=r2 9+y02=r2 9+y02=r2  r=5 x0=5 y0=4  f=(x5)2+(y4)=52A=(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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