AP EAMCET2014MathematicsCircle
The locus of the centre of the circle, which cuts the circle x^2+y^2-20 x+4=0 orthogonally and touches the line x=2 , is
Options
- Ax^2=16 y
- By^2=4 x
- Cy^2=16 x
- Dx^2=4 y
Correct answer
C. y^2=16 x
Step-by-step solution
Let the equation of circle be x^2+y^2+2 g x+2 f y+c=0 where, centre (-g,-f) The centre of given circle x^2+y^2-20 x+4=0 is (10,0) , Condition of two circles cut. aligned 2 (g₁ g₂+f₁ f₂ ) & =c₁+c₂ 2(-g 10+0 (-f) & =c+4 2(-10 g) & =c+4 aligned Also, circle touch the line x=2 . The perpendicular distance from centre to the circle is equal to radius of the circle. aligned & |-g-2| 1 = g^2+f^2-c & (g+2)= g^2+f^2-c & g^2+4+4 g=g^2+f^2-c & f^2-4 g-c-4=0 & f^2-4 g+4+20 g-4=0 & f^2+16 g=0 aligned Hence, the locus of (-g,-f)