Question

Find the equation of the circle touching the line X+2y=1 at (3,-1) and passing through the point (2,1)​

Answer 1

Answer:

the below explanation is the answer

Step-by-step explanation:

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