AP EAMCET2013MathematicsCircle
(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
Options
- A(r^2+b^2+a^2 )^ 1 / 2
- B(r^2+b^2-a^2 )^ 1 / 2
- C(r^2+b^2-a^2 )^ 1 / 3
- D(r^2+b^2+a^2 )^ 1 / 3
Correct answer
B. (r^2+b^2-a^2 )^ 1 / 2
Step-by-step solution
Let the equation of circle whose centre (a, 0) and radius (r) is gathered (x-a)^2+(y-0)^2=r^2 S₁ x^2+a^2-2 a x+y^2-r^2=0 gathered and the equation of circle whose centre (b, 0) and radius R is gathered (x-b)^2+(y-0)^2=R^2 S₂ x^2+b^2-2 b x+y^2-R^2=0 gathered Equation of radical axis is gathered S₁-S₂=0 a^2-b^2+2 b x-2 a x+R^2-r^2=0 R^2=r^2-a^2+b^2-2 b x+2 a x gathered Since, radical axis is y -axis. Therefore, putting x=0 in Eq. (i), we get aligned & R^2=r^2-a^2+b^2-0+0 & R= (r^2+b^2-a^2 )^ 1 / 2 aligned