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# A circle passes through (2,1) and x+2y=1 is a tangent to a circle at (3,-1) find its equation? Please provide with the answer fast!

Arif Hossain
4 years ago
Since x+2y=1 tangent, a diameter will be perpendicular to the line and pass thru (3,-1). The slope of that line is +2, the negative inverse of the line x+2y=1
The eqn of the line thru the center of the circle and thru (3,-1) is:
y= 2x-7
Now, the center of the circle is on that line and equidistant from (3,-1) and (2,1).
so,

sqrt[(x-3)^2 + (y+1)^2] = sqrt[(x-2)^2 + (y-1)^2]

or, x=2y+5/2
Therefore co-ordinate of center is (2y+5/2,y) it's on the line y = 2x-7
so, y=2(2y+5/2)-7
or,y=2/3
The center is (23/6,2/3)
we know the center and poin on circumference of circle so we easily get the equation