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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.

8 years ago

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

equation =(x-5/2)^2+(y-2)^2=(5/2)^2

8 years ago

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