From a point R(5,8) two tangents RP and RQ are drawn to a given circle S=0 whose radius is 5 if circumcenter of triangle PQR is (2,3) then the equation of circle S=0 is ??

Let asume the centre of the circle be (h,k). Therefore its eq be (x-h)²+(y-k)²=25(s=0). Then make the eq of the circle by using the diametric form( using pt R(5,8),centre (h,k))is [(x-h)(x-5)+(y-k)(y-8)=0(eq 1). the circle we get from R and centre of circle S=0, have

centre (2,3). The circle have a radius( under root 34). The eq we get from centre (2,3) and radius ( under root 34)

Is (x-2)²+(y-3)²=34(eq 2). After expanding compare the coefficient of  eq1&2 we get the value of h and k. now,put the value in s=0