AP EAMCET20228 Jul 2022Evening ShiftMathematicsCircleActual
If the segments of the straight lines x+y=6 and x+2 y=4 are two diameters of a circle passing through (6,2) , then the equation of that circle is
Options
- Ax^2+y^2-2 x-4 y-20=0
- Bx^2+y^2+6 x-4 y-68=0
- Cx^2+y^2-16 x+4 y+48=0
- Dx^2+y^2+2 x-10 y-32=0
Correct answer
C. x^2+y^2-16 x+4 y+48=0
Step-by-step solution
L₁ x+y=6 and L₂ x+2 y=4 Point of intersection of lines L₁ and L₂ is centre C of the circle. Solving L₁ and L₂ : x+y=6 x=6-y Then, x+2 y=4 array ll & 6-y+2 y=4 & y=-2 and x=8 array Thus, C: (8,-2) . CP is the radius. C P= (8-6)^2+(2+2)^2 = 4+16 = 20 Equation of circle :(x-8)^2+(y+2)^2=( 20 )^2 x^2+y^2-16 x+4 y+48=0