(a,0) and (b,0) are centres of two circles belonging to a coaxial system of which y -axis is the radical axis. If radius of one of the circles is ' r ', then the radius of the other circle is
Let the equation of circle whose centre (a,0) and radius (r) is (x−a)2+(y−0)2=r2 ⇒S1≡x2+a2−2ax+y2−r2=0 and the equation of circle whose centre (b,0) and radius R is (x−b)2+(y−0)2=R2 ⇒S2≡x2+b2−2bx+y2−R2=0 ∴ Equation of radical axis is S1−S2=0 ⇒a2−b2+2bx−2ax+R2−r2=0 ⇒R2=r2−a2+b2−2bx+2ax ......(i) Since, radical axis is y -axis. Therefore, putting x=0 in Eq. (i), we get R2=r2−a2+b2−0+0 ⇒ R=(r2+b2−a2)1∕2