Q.
A circle has radius 3 units and its centre lies on the line y=x−1. If it passes through the point (7,3), its equation is x2+y2+ax+by+c=0. Here, a,b and c respectively are
Let (h,k) be the centre of the circle. Then, k=h−1. Therefore, the equation of the circle is given by (x−h)2+[y−(h−1)]2=9
Given that, the circle passes through the point (7,3) and hence we get (7−h)2−(3−(h−1))2=9
or (7−h)2+(4−h)2=9
or h2−11h+28=0
which given (h−7)(h−4)=0 h=4 or h=7
Therefore, the required equations of the circles are x2+y2−8x−6y+16=0
or x2+y2−14x−12y+76=0
Comparing above equations with x2+y2+ax+by+c=0
where, a−−8,b−−6,c−16
and a=−14,b=−12,c=76