find the equation of the circle having centre at (3,-4) & touching the line 5x+12y-12=0
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the perpendicular distance from the centre of the circle to the line
(x−3)^2 + (y+4)^2 = 45^22
(x−3)^2 + (y+4)^2 = √45/13
(x−3)^2 + (y+4)^2 = (45/13)^2
(x−3)^2 + (y+4)^2 = √45/√13
plzz mark as brainliest answer !!!
(x−3)^2 + (y+4)^2 = 45^22
(x−3)^2 + (y+4)^2 = √45/13
(x−3)^2 + (y+4)^2 = (45/13)^2
(x−3)^2 + (y+4)^2 = √45/√13
plzz mark as brainliest answer !!!
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