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 .
So here's how I tried to solve it :
Since the circle is passing through the point of intersection of the two circles.We can write the family of circles passing through them by S1 + $\omega $ S2, where $\omega $ is a parameter which produces the infinite circle passing through the point of intersection of the two circles.
$\ x^2+y^2-2x+6y+6 $ + $\omega $( x^2+y^2+2x-6y+6 ) = 0
Simplifying we get :
$\ x^2(1+\omega) $+ $y^2$(1+$\omega $) +x(2$\omega $ - 2) +y(-6$\omega $ - 6) + 6 + 6$\omega = 0 $
Since this is a circle it has its centre at $\ (1-\omega),(3\omega + 3) $
Now it is given in the question that a circle from this family is orthogonal to the given circle:
Note: Here I mention first circle as the one you wrote in the diameter form.
$\ r_1^2 + r_2^2 = d^2 $ where $\ r_1 $ is the radius of the first circle and $\ r_2 $ is the radius of the second circle, d is the distance between the two centres of the circles.
First circle has centre (2,5) and radius $\sqrt 10 $
Putting in the values in the formula we get :
$\ 10 $ + (1 - $\omega $)^2 + (3 $\omega $ + 3 - 5)^2 - 6 - 6$\omega $ = (1- $\omega $ - 2)^2 + (3 $\omega $ + 3 - 5) ^2:
Simplifying it all we get $\omega $ = -1/4.
Which gives us the circle when we put the value of $\omega $ in the family of circle to get the required circle.
We get the radius (root of 77 )/4