Geometry and Combinatorics

If you have a circle with radius r, for the point to be closer to the circumference than the center it needs to be further than half of the radius away from the center.

The total area of the circle is: $\pi r^2$ The area of the circle where the point would be closer to the center (where the point would be within half the radius) is : $\pi (\frac{r}{2})^2 = \frac{\pi r^2}{4}$ Therefore the area of the circle where the point is further away is: $\pi r^2 - \frac{\pi r^2}{4} = \frac{3\pi r^2}{4}$

The probability that the point is in the region where its closer to the circumference is the area of that region divided by the total area which is: $\frac{\frac{3\pi r^2}{4}}{\pi r^2} = \frac{3}{4}$

and so $a+b = 3+4 = \boxed{7}$

Let the area of a circle be A

Now, all the points closer to the centre will lie inside the circle of area $\frac{A}{4}$

Therefore, all the points closer to the circumference lie in the area of $A - \frac{A}{4} = \frac{3 \times A}{4}$

$P = \frac{3}{4}$

3 + 4 = 7

For a given point inside the circle, if it has to be closer to the circumference than the centre, its distance from the centre should be $> \dfrac{r}{2}$

Check this image

If the point lies in the blue area, the point will be closer to the circumference than the centre. The area of this blue portion is given by

$\pi r^{2} - \pi \left( \dfrac{ r}{2} \right)^{2}= \dfrac{3 \pi r^{2}}{4}$

The total area (blue + white area) is $\pi r^{2}$

Therefore, the probability that a random point lies in the blue area is

$\dfrac{\dfrac{3 \pi r^{2}}{4}}{\pi r^{2}}$ or $\dfrac{3 \pi r^{2}}{4 \pi r^{2}}$

Simplifying this expression and cancelling like terms we get

$\dfrac{3}{4}$ giving us the answer $\boxed{7}$

This question is easier to answer with a diagram, really. But bear with me...

WLOG, let the radius of the circle be equal to 2.

In order for a random point in the circle to be closer to the circumference that the center, it must be within the annulus (click this for more info. It's really interesting!) that is formed by the concentric circle of radius 1 and the circle itself of radius 2.

The problem is now reduced to finding the area of the annulus. This is calculated quite simply, by subtracting the area of the smaller circle from the area of the larger one, resulting in an area of 3π (4π - π).

The probability that a point is located within this region is equal to 3π/4π or 3/4.

Therefore, the value of a + b is $\boxed{7}$ .

Let the radius of the circle be $r$ and the center be $O$ . For a point $P$ inside the circle to be closer to the circumference of the circle than the center, we must have $OP>\frac{r}{2}$ . For this to happen, $P$ must lie outside the circle centered at $O$ with radius $\frac{r}{2}$ . We have that the probability of $P$ lying outside this circle is $1-\frac{\pi*(\frac{r}{2})^2}{\pi*r^2}=1-\frac{1}{4}=\frac{3}{4}$ Hence, the answer is $3+4=7$ .

Just "visualize" a donut :D Probability (P) = Area of donut / Area of circle

set width of donut = 0.5; set width of circle = 1

Then, P = pi(1)^2 - pi(0.5)^2 / pi(1)^2; P= 3/4; 3+4 = 7

For a point to be closer to the circumference than the center, the following must be met: $R>r>R/2$ Thus the area available is that of the whole circle minus the circle with radius R/2. This gives us $\frac{\pi R^{2} - \pi (R/2)^{2}}{\pi R^{2}}=1-\frac{1}{4} = \frac{3}{4}$ Thus $a+b=\boxed{7}$

let OA=R is the radius of the O centered circle, take a point B on line OA such that OB=BA, let OB=r, so r=R/2.

a random point inside the circle closer to the circumference could be anywhere in the area of (πR^2-πr^2)

thus probability is, (πR^2-πr^2)/(πR^2) = 1 - 1/4 = 3/4. 3+4=7

consider a circle of radius 2r. let a concentric circle of radius r. logically a point lying in the smaller circle is closer to center. therefore (total area-area of smaller circle)/total area =reqd. probability

consider a smaller circle of raddi a/2 so it will be closer to centre than to....circumference..........so probabilty of this=smaller cirle area /bigger circle area.......=1/4.......so probability reqd=1-1/4=3/4.............