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AP EAMCET202021 Sep 2020Morning ShiftMathematicsCircleActual

Find the equation of circle having normals ((x-1)(y-2)=0 ) and a tangent (3 x+4 y=6 ) ?

Options

  1. A((x-1)^2+(y-2)^2=1 )
  2. B((x-2)^2+(y-1)^2=1 )
  3. C((x+1)^2+(y+2)^2=1 )
  4. D((x+2)^2+(y+1)^2=1 )

Correct answer

A. ((x-1)^2+(y-2)^2=1 )

Step-by-step solution

The equation of normals to the circle are (x-1=0 ) and (y-2=0 ), so centre of the circle is ((1,2) ) and since (3 x+4 y=6 ) is the tangent to the circle so radius (r= 3+8-6 3^2+4^2 =1 ) ( ) Equation of required circle is ((x-1)^2+(y-2)^2=1 ) Hence, option (a) is correct.

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