Applications of Circles

Applications of circles

We have looked at the general equation of a circle, $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2, which is centred at the point $\left(h,k\right)$(h,k) and has a radius of $r$r. A special case of this equation occurs when the circle is centred at the origin. In this case, $h$h and $k$k are both $0$0 and so the equation takes the form $x^2+y^2=r^2$x2+y2=r2.

We can make use of our knowledge of circles and their equations to solve various problems involving circles. Let's take a look at a few properties.

• The equation of a circle describes all of the points that lie exactly a distance of $r$r away from the centre point $\left(h,k\right)$(h,k). Any points which are closer than $r$r units to the centre lie inside the circle, and any points that are further than $r$r units from the centre lie outside the circle.

• Two points which lie on opposite ends of a diameter of a circle have the centre of the circle as their midpoint. 

• The area of a circle of radius $r$r is given by $A=\pi r^2$A=πr2. Similarly, the length of the circumference of a circle of radius $r$r is given by $C=2\pi r$C=2πr.
• A circle can be inscribed in a square as in the image below. This square is the smallest possible square that can fit around the circle, which touches at four points and has side lengths equal to the length of the diameter of the circle.

 

Let's go through an example now.

Worked example

A circle of radius $3$3 units is centred at the point $\left(12,5\right)$(12,5). What is the shortest distance from the origin to the circle?

Think: We can start by drawing a diagram of the situation, to easily see what's going on.

From the diagram, we can see that the closest point to the origin is between the centre and the origin. In fact, if we draw a line from the origin to the centre of the circle it will intersect the circle at the closest point.

This means that the shortest distance from the origin to the circle will be equal to the distance between the origin and the centre, minus the radius of the circle.

Do: We know that the radius of the circle is $3$3 units, and we can use the distance formula to find the distance between the origin and the centre:

$d$d	$=$=	$\sqrt{\left(12-0\right)^2+\left(5-0\right)^2}$√(12−0)2+(5−0)2
	$=$=	$\sqrt{12^2+5^2}$√122+52
	$=$=	$\sqrt{144+25}$√144+25
	$=$=	$\sqrt{169}$√169
	$=$=	$13$13

Subtracting the radius from this, we can see that the shortest distance from the origin to the circle is $13-3=10$13−3=10 units.

 

Practice questions

Question 1

Question 2

Question 3

Applications of semi-circles

We know that the general equation of a circle is $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(x−h)2+(y−k)2=r2, which is centred at the point $\left(h,k\right)$(h,k) and has a radius of $r$r.

From this we can find the equations of the upper- and lower-half semicircles by solving for $y$y.

Upper: $y=k+\sqrt{r^2-\left(x-h\right)^2}$y=k+√r2−(x−h)2

Lower: $y=k-\sqrt{r^2-\left(x-h\right)^2}$y=k−√r2−(x−h)2

Similarly, the equations of the left- and right-half semicircles can be found by solving for $x$x.

Right: $x=h+\sqrt{r^2-\left(y-k\right)^2}$x=h+√r2−(y−k)2

Left: $x=h-\sqrt{r^2-\left(y-k\right)^2}$x=h−√r2−(y−k)2

In the special case that the semicircles are centred at the origin, so $h=0$h=0 and $k=0$k=0, these equations simplify to $y=\pm\sqrt{r^2-x^2}$y=±√r2−x2 and $x=\pm\sqrt{r^2-y^2}$x=±√r2−y2.

 

We can make use of these equations, and our knowledge of circles, to solve various problems involving semicircles. Let's take a look at a few properties.

• The equation of a semicircle describes a set of points that lie exactly a distance of $r$r away from the centre point $\left(h,k\right)$(h,k), and are to one side of the centre. Any points on the same side which are closer than $r$r units to the centre lie inside the semicircle, and any points that are further than $r$r units from the centre or on the other side lie outside the semicircle.

• Any diameter divides a circle into two semicircles. The two endpoints of the diameter have the centre of the circle as their midpoint.

• The area of a circle of radius $r$r is given by $A=\pi r^2$A=πr2. Similarly, the length of the circumference of a circle of radius $r$r is given by $C=2\pi r$C=2πr.
  Correspondingly, the area of a semicircle of radius $r$r is given by $A=\frac{1}{2}\pi r^2$A=12​πr2. The length of the curved part of the perimeter is given by $L=\pi r$L=πr, and the total perimeter is given by $P=\pi r+2r$P=πr+2r.

 

Practice questions

Question 4

Question 5

Question 6