MHT CET20225 Aug 2022Morning ShiftMathematicsCircleActual
The equation of the circle whose centre lies on the line x-4 y=1 and which passes through the points (3,7) and (5,5) is
Options
- Ax^2+y^2+6 x-2 y+90=0
- Bx^2+y^2+6 x+2 y+90=0
- Cx^2+y^2+6 x+2 y-90=0
- Dx^2+y^2-6 x+2 y-90=0
Correct answer
C. x^2+y^2+6 x+2 y-90=0
Step-by-step solution
Let the equation of the circle be x^2+y^2+2 g x+2 f y+c=0 its centre lies on x -4 y =1 - g -4(- f )=1 it passes through (3,7) 3^2+7^2+2 ~g 3+2 f 7+ c =0 Also it passes through (5,5) 5^2+5^2+2 g 5+2 f 5+c=0 from (iii) - (ii) 4 g-4 f=8 g-f=2 aligned & f =1 & g =3 & c =-90 aligned So, the equation of required circle is x^2+y^2+6 x+2 y-90=0