Circles and triangles 

In a nutshell

The circumcircle of a triangle is a circle that can be drawn to pass through the three vertices of the triangle. The centre of a circumcircle is the circumcentre and is the point where the any two perpendicular bisectors of each side intersect. For a right-angled triangle, the hypotenuse of the triangle is the diameter of the circumcircle. As perpendicular bisectors of chords intersect at the centre of the circle, using any three points on the circumference of the circle, the centre can be found by finding the intersection point of perpendicular bisectors. 

Circumcircles 

​The circumcircle of a triangle is a circle that can be drawn to pass through the three vertices of the triangle.  ​

​The centre of a circumcircle is the circumcentre and is the point where any two perpendicular bisectors of each side intersect.​

For a right-angled triangle, the hypotenuse (the longest side opposite the right-angle) is the diameter of the circumcircle. 

If $$\angle{ACB}$$ is $${90}\degree$$ then the line segment $$AB$$ is the diameter of the circle. ​

Finding the centre of a circle with three points

Using three points on the circumference of a circle, two chords can be drawn. Using the two chords of a circle, the perpendicular bisectors can be found which intersect at the centre of the circle. To find the centre of the circle, find the equations of the perpendicular bisectors then the point at which they intersect. ​

Example 2

The points $$A(-6,3)$$, $$B(-3,12)$$ and $$C(0,3)$$​ lie on the circumference of the circle. 

a) Find the equation of the perpendicular bisector $$AB$$.​

First, find the gradient of the line segment $$AB$$, using $$m= \dfrac{y_2-y_1}{x_2 -x_1}$$.

$$m= \dfrac{3-12}{-6 --3}=3$$​

So the gradient of its perpendicular bisector is $$-\dfrac{1}{3}$$.

Next, find the midpoint of AB using $${\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}$$.

$${\bigg(\dfrac {(-6 + -3)}{2} , \dfrac {(3 + 12)}{2}\bigg)}\\{\bigg(\dfrac {-9}{2} , \dfrac {15}{2}\bigg)}$$

​

Substitute the coordinates of the midpoint of$$AB$$​ into $$\\\\y-y_1=-\dfrac{1}{3}(x-x_1)$$ to find the equation of $$AB$$.

$$\begin{aligned}y-\dfrac {15}{2}&=-\dfrac {1}{3}(x--\dfrac {9}{2})\\y-\dfrac {15}{2}&=-\dfrac {1}{3}x-\dfrac {3}{2}\end{aligned}\\ \ \ \space\underline{y=-\dfrac{1}{3}x+6}$$

​​​

b) Find the equation of the perpendicular bisector $$BC$$.

First, find the gradient of the line segment $$BC$$​, using $$m= \dfrac{y_2-y_1}{x_2 -x_1}$$.

$$m= \dfrac{12-3}{-3 -0}=-3$$​​

​​

So the gradient of its perpendicular bisector is $$\dfrac{1}{3}$$.

Next, find the midpoint of $$BC$$ using $${\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}$$.

$${\bigg(\dfrac {(-3 + 0)}{2} , \dfrac {(12 + 3)}{2}\bigg)}\\{\bigg(\dfrac {-3 }{2} , \dfrac {15}{2}\bigg)}$$

​​​

Substitute the coordinates of the midpoint of $$BC$$ into $$\\y-y_1=\dfrac{1}{3}(x-x_1)$$ to find the equation of $$BC$$.

$$\begin{aligned}y-\dfrac{15}{2}&=\dfrac{1}{3}(x--\dfrac{3}{2})\\y-\dfrac{15}{2}&=\dfrac{1}{3}x+\dfrac{1}{2}\end{aligned}\\ \space\space\space\underline{y=\dfrac{1}{3}x+8}$$

​​​

c) Find the centre of the circumcircle passing through points $$ABC$$.​

Find the point of intersection between perpendicular bisectors $$AB$$ and $$BC$$.

​Make $$x$$ the subject of $$BC$$:

$$x=3y-24$$

Substitute the equation for $$BC$$ into the equation for $$AB$$ and solve for $$y$$.

$$\begin{aligned}y&=-\dfrac{1}{3}(3y-24)+6 \\3y&=-(3y-24)+18\\3y&=-3y+24+18\\6y&=42\\y&=7\end{aligned}$$​​

​​​​​​

Substitute $$y=7$$ into $$x=3y-24$$ to solve for $$x$$.

$$\begin{aligned}x&=3(7)-24\\x&=-3\end{aligned}$$​​

Therefore, the centre of the circumcircle is $$\underline{(-3,7)}$$​​

d) Find the equation of the circumcircle. 

To find the radius of the circle find the distance between the centre $$(-3,7)$$​ and a point on the circumference of the circle. 

 $$\begin{aligned}d&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\&=\sqrt{(-3--3)^2+(12-7)^2}\\&=\sqrt{0+5^2}\\&=5\end{aligned}$$,

Therefore, the equation of the circumcircle is $$\underline{(x+3)^2+(y-7)^2=25}$$

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