The equation of a circle which cuts the three circles $x^2+y^2-3x-6y+14=0,x^2+y^2-x-4y+8=0$

$x^2+y^2+2x-6y+9=0$ orthogonally, is

detailed solution

Correct option is A

The circle having centre at the radical centre of three given circles and radius equal to the length of the tangent from it to any one of three circles cuts all the three circles orthogonally. The given circles are x2+y2−3x−6y+14=0           …(i)x2+y2−x−4y+8=0              …(ii)x2+y2+2x−6y+9=0          …(iii)The radical axes of (i), (ii) and (ii), (iii) are respectively x+y−3=0                 …(iv)and, 3x−2y+1=0           …(v)Solving (iv) and (v), we get x=1,y=2Thus, the coordinates of the radical centre are (1, 2). The length of the tangent from (1, 2) to circle (i) is given by r=1+4−3−12+14=2Hence, the required circle is (x−1)2+(y−2)2=22⇒x2+y2−2x−4y+1=0