Q57 / 150 · MHT CET 2011 · Mathematics
The equation of a circle which has a tangent $3 x+4 y=6$ and two normals given by $(x-1)(y-2)=0$ is
- A$(x-3)^{2}+(y-4)^{2}=5^{2}$
- B$x^{2}+y^{2}-4 x-2 y+4=0$
- Correct answerC$x^{2}+y^{2}-2 x-4 y+4=0$
- D$x^{2}+y^{2}-2 x-4 y+5=0$
Solution
$(x-1)(y-2)=0$
$$
\begin{array}{ll}
\Rightarrow \quad x-1=0 \text { and } y-2=0 \\
\therefore \quad \text { Radius }=\frac{3(1)+4(2)-6}{\sqrt{9+16}} \\
=\frac{5}{5}=1
\end{array}
$$
$\therefore$ Equation of the circle is
$$
(x-1)^{2}+(y-2)^{2}=1
$$
or $\quad x^{2}+y^{2}-2 x-4 y+4=0$
Hence, the correct answer is option: 'C'.