Question

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

• A. $(x-1)^{2}+(y-2)^{2}=1$
• B. $(x-2)^{2}+(y-1)^{2}=1$
• C. $(x+1)^{2}+(y+2)^{2}=1$
• D. $(x+2)^{2}+(y+1)^{2}=1$

Answer 1

Answer: (A)

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 $r=\frac{3+8-6}{\sqrt{3^{2}+4^{2}}}=1$
$\therefore $ Equation of required circle is
$(x-1)^{2}+(y-2)^{2}=1$