Question

The number of points P on circle $(x-3)(x-4)+(y-1)(y-2)=0$ such that area of $△PAB$ is 1/2 sq units where A $(3,2)$ and B$(4,1)$ is:

• A. four
• B. sixteen
• C. two
• D. none of these

Answer 1

The correct option is A four

Let $P$ be $(h,k)$

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}h& k& 1\\ 3& 2& 1\\ 4& 1& 1\end{array}\right|=\frac{1}{2}$

$\Rightarrow \left|h\right(2-1)-k(3-4)+1(3-8\left)\right|=1$

Taking $+$ sign ($\because$ mod will give both $+$ and $-$)

$\Rightarrow h+k-5=1$

$\Rightarrow h+k=6\to \left(1\right)$

$\because P$ is a point on circle.

$\Rightarrow (h-3)(h-4)+(k-1)(k-2)=0$

$\Rightarrow h^2-7h+12+k^2-3k+2=0$

$\Rightarrow h^2+k^2-7h-3k+14=0\to \left(2\right)$

Using $\left(1\right)$ and putting $k=6-h$

$\Rightarrow h^2+(6-h{)}^2-7h-3(6-h)+14=0$

$h^2+36+h^2-12h-7h-18+3h+14=0$

$\Rightarrow 2h^2-16h+32=0$

$\Rightarrow h^2-8h+16=0$

$\Rightarrow (h-4{)}^2=0$

$\Rightarrow h=4$

$k=6-h=6-4=2$

$\therefore (4,2)$ is the only point on circle which satisfies the given condition on ${}^{\prime }{+}^{\prime }$ sign.

On taking ${}^{\prime }{-}^{\prime }$ sign:

$h+k-5=-1$

$\Rightarrow h+k=4$

Putting in $\left(2\right)$

$\Rightarrow h^2+(4-h{)}^2+14-7h-3(4-h)=0$

$\Rightarrow h^2+16+h^2-8h+14-7h-12+3h=0$

$\Rightarrow 2h^2-12h+18=0$

$\Rightarrow h^2-6h+9=0$

$\Rightarrow (h-3{)}^2=0$

$\Rightarrow h=3$

$k=4-h=1$

$\therefore (3,1)$ is the point on circle which satisfies the given condition on taking ${}^{\prime }{-}^{\prime }$ sign.

$\therefore$ Two points.

$\therefore C)$ Answer.