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A circle having its centre in the first quadrant touches the y axis at point (0,2) and passes through the point (1,0) . find the equation of circle.
look since circle touches y axis at (0,2) hence perpendicular from this point passes through centre as y axis is tangent to it.. hence coordinate of circle would be (a,2) now its disatnce from (0,2)and (1,0) is equal as its the radius itself hence √(a-0)^2+(2-2)^2=√(a-1)^2+(2-0)^2 a^2=a^2+1-2a+4 a=5/2 hence radius =5/2 equation =(x-5/2)^2+(y-2)^2=(5/2)^2
look
since circle touches y axis at (0,2)
hence perpendicular from this point passes through centre as y axis is tangent to it..
hence coordinate of circle would be (a,2)
now its disatnce from (0,2)and (1,0) is equal as its the radius itself
hence √(a-0)^2+(2-2)^2=√(a-1)^2+(2-0)^2
a^2=a^2+1-2a+4
a=5/2
hence radius =5/2
equation =(x-5/2)^2+(y-2)^2=(5/2)^2
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