Question

Diameter $AB$ of a circle with centre $O$ is $10$ units. $C$ is a point $4$ units from $A$, and on $AB$. $D$ is a point $4$ units from $B$, and on $AB$. $P$ is any point on the circle. Then the broken-line path from $C$ to $P$ to $D$:

• A. has the same length for all positions of $P$
• B. exceeds $10$ units for all positions of $P$
• C. cannot exceed $10$ units
• D. is shortest when $CPD$ is a right triangle
• E. is longest when $P$ is equidistant from $C$ and $D$

Answer 1

The correct option is B exceeds $10$ units for all positions of $P$

If $Q$ is the foot of the perpendicular from $P$ to $\overline{AB}$ and let $\overline{QB}=x$
Then, ${\left(PQ\right)}^2=x(10-x)\overline{CP}=\sqrt{x(10-x)+{(6-x)}^2}\overline{DP}=\sqrt{x(10-x)+{(4-x)}^2}\overline{CP}+\overline{DP}=\sqrt{36-2x}+\sqrt{16+2x}$
We can see that in the equation $\sqrt{36-2x}+\sqrt{16+2x}$ is maximum only when $x=5.$ Even the ellipse through $A$ and $B$ with $C$ and $D$ as foci is the locus of points $P^{\prime }$ where $\overline{C{P}^{\prime }}+\overline{P^{\prime }O}=10,$ and as the ellipse lies entirely in side the given circle, we see that $\overline{CP}+\overline{DP}\ge 10$ for all position of $P.$