Question

The normal at the point $(2,3)$ to the circle $x^2+y^2-2x-4y+3=0$ intersects the circle $x^2+y^2=1$ at the points $P$ and $Q$. The area of the circle with $PQ$ as diameter is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{2}$

The normal to the circle $x^2+y^2-2x-4y+3=0$ at $(2,3)$ would pass through the center of the circle also i.e. $(1,2)$

Equation of the normal to the circle is
$y-3=\frac{2-3}{1-2}(x-2)\Rightarrow y=x+1\cdots \left(1\right)$
$y=x+1$ intersection with the circle $x^2+y^2=1$,
$x^2+(x+1{)}^2=1\Rightarrow 2x^2+2x=0\Rightarrow x=0,-1$
So,
$P(0,1)$ and $Q(-1,0)$ are the points and
$\Rightarrow PQ=\sqrt{2}$
Hence, the area of the circle with $PQ$ as diameter will be
$=\pi \frac{d^2}{4}=\frac{\pi }{2}$