Tangents are drawn to circle $x^2+y^2-6x-4y-11=0$ from point $P(1,8)$ touching circle at $A$ and $B$. Let there be a circle whose radius passes through point of intersection of circles $x^2+y^2-2x-6y+6=0$ and $x^2+y^2+2x-6y+6=0$ and intersect the circumcircle of $PAB$ orthogonally. Find minimum radius of such a circle.

My attempt Circumcircle would be $(x-1)(x-3)+(y-2)(y-8)=0$ and centre would lie on radical axis of the given circles ie $x=0$ . therefore equation of circle will be $x^2+y^2+2fx+c=0$. Applying orthogonality condition , $c=-19-10f$ .

I am unable to find minimum radius .

Please suggest any ways to minimise the radius or any better solution if possible .

  • $\begingroup$ According to relations you derived, $r^2 = f^2 - c = f^2+10f+19$, meaning that minimum $r$ will be zero! $\endgroup$ – samjoe Sep 25 '17 at 7:16
  • $\begingroup$ @samjoe what should be circumcircle according to you ? $\endgroup$ – Seema Prasad Sep 25 '17 at 7:28
  • $\begingroup$ Sorry I checked again. Its correct. $\endgroup$ – samjoe Sep 25 '17 at 7:43
  • $\begingroup$ What do you mean by "whose radius passes through point of intersection of circles" ? Let I and J be these points. Do you mean that there are two choices of circles : one with center I and passing by J, the other with center J passing by I. Or do you mean that we are only interested by length IJ and that the center of the circle can be elsewhere ? $\endgroup$ – Jean Marie Sep 25 '17 at 12:27
  • $\begingroup$ The meaning of "a circle whose radius passes through point of intersection of ..." is not clear at all. Do they mean the line to which the radius belongs? Or what? $\endgroup$ – Aretino Sep 25 '17 at 12:47

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