Question

find the equation of the circle passing through (6,3) and touching the coordinate axes.​

Answer 1

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Answer 2

Answer:

Step-by-step explanation

as it is touching coordinate axis

g^2=c

f^2=c

that means g=f=k

let equation of circle be

x^2+y^2+2gx+2fy+c  =0

x^2+y^2+2kx+2ky+k^2=0

as 6.3 passes through circle it satisfies

36+9+12k+6k+k^2 = 0

45+k^2+18k = 0

k=-18+minus 324-180 / 2

k=-18+minus 144 / 2

k=-18+minus 12 / 2

k = -3 or -15

therefore

g=f=-3 or g=f=-15

c = 9 or 225

equation of the circle

x^2+y^2-6x-6y+9 = 0

x^2+y^2-30x-30y+225=0