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MHT CET20244 May 2024Evening ShiftMathematicsCircleActual

The equation of the circle which passes through the centre of the circle x^2+y^2+8 x+10 y-7=0 and concentric which the circle 2 x^2+2 y^2-8 x-12 y-9=0 is

Options

  1. Ax^2+y^2-4 x-6 y+77=0
  2. Bx^2+y^2-4 x-6 y-89=0
  3. Cx^2+y^2-4 x-6 y+97=0
  4. Dx^2+y^2-4 x-6 y-87=0

Correct answer

D. x^2+y^2-4 x-6 y-87=0

Step-by-step solution

Required circle is concentric with aligned & 2 x^2+2 y^2-8 x-12 y-9=0 & x^2+y^2-4 x-6 y- 9 2 =0 aligned Centre is (2,3) Also, it passes through centre of x^2+y^2+8 x+10 y-7=0 Centre is (-4,-5) Radius = (-4-2)^2+(-5-3)^2 =10 Equation of required circle is aligned & (x-2)^2+(y-3)^2=10^2 & x^2+y^2-4 x-6 y-87=0 aligned

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