MHT CET20244 May 2024Morning ShiftMathematicsCircleActual

The equation of the çircle which has its centre at the point (3,4) and touches the line 5 x+12 y-11=0 is

Options

Ax^2+y^2-6 x-8 y+9=0
Bx^2+y^2-6 x-8 y+25=0
Cx^2+y^2-6 x-8 y-9=0
Dx^2+y^2-6 x-8 y-25=0

Correct answer

A. x^2+y^2-6 x-8 y+9=0

Step-by-step solution

aligned & Radius = Distance of a point (3,4) form & 5 x+12 y-11=0 & = | 5(3)+12(4)-11 25+144 | & = | 15+48-11 169 | & = 52 13 & =4 aligned Required equation is aligned & (x-3)^2+(y-4)^2=(4)^2 & x^2+y^2-6 x-8 y+9=0 aligned

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