Question

The equation of a circle which has a tangent $ 3x + 4y = 6 $ and two normals given by $ (x - 1 ) (y - 2 ) = 0 $ is

• A. $ (x - 3)^2 + (y - 4)^2 = 5^2 $
• B. $ x^2 + y^2 - 4x - 2y + 4 = 0 $
• C. $ x^2 + y^2 - 2x - 4y + 4 = 0 $
• D. $ x^2 + y^2 - 2x - 4y + 5 = 0 $

Answer 1

Answer: (C)

$(x-1)(y-2)=0$
$\Rightarrow x-1=0$ and $y-2=0$
$\therefore $ Radius $=\frac{3(1)+4(2)-6}{\sqrt{9+16}}$
$=\frac{5}{5}=1$
$\therefore $ Equation of the circle is
$(x-1)^{2} +(y-2)^{2}=1$
or $x^{2}+y^{2}-2 x-4 y+4=0$