Q55 / 170 · Manipal (MET) 2019 · Mathematics
The normal at the point $(3,4)$ on a circle cuts the circle at the point $(-1,-2)$. Then the equation of the circle is
- A$x^2+y^2+2 x-2 y-13=0$
- Correct answerB$x^2+y^2-2 x-2 y-11=0$
- C$x^2+y^2-2 x+2 y+12=0$
- D$x^2+y^2-2 x-2 y+14=0$
Solution
Given that normal cuts the circle at points $A(3,4)$ and $B(-1,-2)$. As we know, normal line is always passing through the centre of the circle and $A B$ is the diameter of the circle. Therefore, $A$ and $B$ are the end points of a diameter.
$\therefore$ Required equation of circle is
$\begin{aligned}
& (x-3)(x+1)+(y-4)(y+2)=0 \\
\Rightarrow & x^2-2 x-3+y^2-2 y-8=0 \\
\Rightarrow & x^2+y^2-2 x-2 y-11=0
\end{aligned}$
Hence, the correct answer is option: 'B'.