P is a point inside a circle and A is a point on the circumference. The minimum distance between A and P is 2 cm and the maximum distance between A and P is 8 cm. Find the radius of the circle. I think the radius must be greater than 4 cm. But next how to proceed ??
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Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$? – Shubham Johri Jan 24 '19 at 11:09
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Are you familiar with triangular inequality? – dfnu Jan 24 '19 at 11:16
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@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm. – Subham Sen Jan 24 '19 at 12:33
2 Answers
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You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example, $$A'P+OP = r,$$ where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$, $$AP+OP \geq r,$$ so that $$AP \geq A'P.$$ So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus $$A'P=2\ \mbox{cm}.$$ Similarly, you can express $A''P$ as $$A''P = r + OP,$$ whereas, again for triangular inequality on $AOP$, $$AP \leq r + OP,$$ which yields $$ AP \leq A''P.$$ In conclusion $A''P = 8$ cm and the radius is $r = \frac{A'P + A''P}{2}=5$ cm.
$\hskip1.5in$
dfnu
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Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ?? – Subham Sen Jan 24 '19 at 12:22
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@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius. – dfnu Jan 24 '19 at 13:04
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b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ?? – Subham Sen Jan 24 '19 at 18:56
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@SubhamSen question b> must be dealt first. What I meant is that if the point $A$ is not taken on the diameter passing through point $P$, then its distance from $P$ will be greater than the distance of $A'$ from $P$, $A'$ being exactly the point intersected on the circle by the diameter passing through $P$. And this is just a rephrasing of what was demonstrated via triangular inequality. – dfnu Jan 24 '19 at 21:05
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The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
