AP EAMCET201920 Apr 2019Morning ShiftMathematicsCircleActual
The equation of the circle whose radius is 3 and which touches internally the circle (x^2+y^2-4 x-6 y-12=0 ) at the point ((-1,-1) ) is
Options
- A(5 x^2+5 y^2+9 x-6 y-7=0 )
- B(5 x^2+5 y^2-8 x-14 y-32=0 )
- C(5 x^2+5 y^2-6 x+8 y-8=0 )
- D(5 x^2+5 y^2+6 x-8 y-12=0 )
Correct answer
B. (5 x^2+5 y^2-8 x-14 y-32=0 )
Step-by-step solution
Equation of given circle is (x^2+y^2-4 x-6 y-12=0 ) having centre (C₁(2,3) ) and radius (r₁= 4+9+12 =5 ). Let the required circle having centre (C₂(h, k) ) and radius is given as 3 touches the given circle at (A(-1,-1) ). The point (A(-1,-1) ) divides the line joining the centres (C₁(2,3) ) and (C₂(h, k) ) externally in (5: 3 ) so ( aligned & (-1,-1)= ( 5 h-3(2) 5-3 , 5 k-3(3) 5-3 ) & (-1,-1)= ( 5 h-6 2 , 5 k-9 2 ) & 5 h-6=-2 and 5 k-9=-2 & h= 4 5 and k= 7 5 aligned ) so equation of required circle is ( aligned & (