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=(x0−6/2,0)  (x−x0)2+(y−y0)2=r2  (0−x0)2+(4−y0)2=r2 (x0+3−x0)2+(0−y0)2=r2 (x0−3−x0)2+(0−y0)2=r2  x02+(4−y0)2=r2 9+y02=r2 9+y02=r2  r=5 x0=5 y0=4  f=(x−5)2+(y−4)=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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