A circle passing through the point P( , ) in the first quadrant touches the two coordinate…

Question

A circle passing through the point $P(\alpha ,\beta )$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$. The point $P$ is above the line $AB$. The point $Q$ on the line segment $AB$ is the foot of perpendicular from $P$ on $AB$. If $PQ$ is equal to 11 units, then the value of $\alpha \beta $ is
 \_\_\_\_\_\_\_\_\_\_\_.

Answer: 121

Accepted Answer

Let equation of circle is $(x-a{)}^{2}+(y-a{)}^{2}={a}^{2}$

Since, (i) passes through $P(\alpha ,\beta )$

$	herefore (\alpha -a{)}^{2}+(\beta -a{)}^{2}={a}^{2}$

$\begin{gathered} \Rightarrow {\alpha }^{2}+{\beta }^{2}-2\alpha a-2\beta a+{a}^{2}=0.........(i) \end{gathered}$

Equation of $AB=\frac{x}{a}+\frac{y}{a}=1$

$\Rightarrow x+y=a$ .........(ii)

Let $Q(p,q)$ be the foot of the perpendicular from $P$ to line (iii)

$\begin{gathered} 	herefore \frac{p-\alpha }{1}=\frac{q-\beta }{1}=\frac{-(\alpha +\beta -\alpha )}{(1{)}^{2}+(1{)}^{2}} \ \Rightarrow \frac{p-\alpha }{1}=\frac{q-\beta }{1}=\frac{-(\alpha +\beta -a)}{2} \ \Rightarrow p-\alpha =\frac{-(\alpha +\beta -a)}{2} \ q-\beta =\frac{-(\alpha +\beta -a)}{2} \end{gathered}$

Now,

$\begin{gathered} P{Q}^{2}=(p-\alpha {)}^{2}+(q-\beta {)}^{2} \ =\frac{1}{4}(\alpha +\beta -a{)}^{2}+\frac{1}{4}(\alpha +\beta -a{)}^{2} \end{gathered}$

$\Rightarrow (11{)}^{2}=\frac{1}{2}(\alpha +\beta -\alpha {)}^{2}$

$\begin{gathered} \Rightarrow {\alpha }^{2}+{\beta }^{2}+{a}^{2}+2\alpha \beta -2\beta a-2\alpha a=242 \ \Rightarrow 2\alpha \beta =242 \ \Rightarrow \alpha \beta =121 \end{gathered}$

Solutions · 1