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

Question

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

Answer: 121

Accepted Answer

Circle: $(x-a)^2+(y-a)^2=a^2$ (centre $(a,a)$, radius $a$). Through $P$: $\alpha^2+\beta^2-2a\alpha-2a\beta+a^2=0$.

Line $AB$: $x+y=a$. Distance from $P$ to $AB$:
$$PQ=\frac{|\alpha+\beta-a|}{\sqrt2}\ \Longrightarrow\ PQ^2=\frac{(\alpha+\beta-a)^2}2.$$

Expanding, $(\alpha+\beta-a)^2=(\alpha^2+\beta^2-2a\alpha-2a\beta+a^2)+2\alpha\beta=2\alpha\beta$ (using the circle relation). So $PQ^2=\alpha\beta$, giving $\alpha\beta=121.$

Solutions · 1