Tangents drawn from the point P (1, 8) to the circle x^2 + y^2 − 6x − 4y − 11 = 0 touch the circle at the points
A and B. The equation of the circumcircle of the triangle PAB is

• Write d eqn of a general line thru (1,8) and (h,k) and let this line's distance thru the center of the circle be the radius [ centre is (-g,-f) and radius is WHOLE ROOT OF g^2+f^2-c of any circle of the form x^2+y^2+2gx+2fy+c=0]
  Thus , we obtain the eqn of the lines passing thru d vertices of d triancle. Find any point (l,m) equidistant from the vertices of this triangle, that's d circumcentre and d eqn of d circumcircle will be of the form (x-l)^2+(y-m)^2=(rad)^2
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