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Tutor: Labib

Summary

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.  ​

Maths; Circles; KS5 Year 12; Circles and triangles

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

Maths; Circles; KS5 Year 12; Circles and triangles

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

Maths; Circles; KS5 Year 12; Circles and triangles

If ∠ACB\angle{ACB}∠ACB is 90°{90}\degree90° then the line segment ABABAB 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)A(-6,3)A(−6,3), B(−3,12)B(-3,12)B(−3,12) and C(0,3)C(0,3)C(0,3)​ lie on the circumference of the circle. 


a) Find the equation of the perpendicular bisector ABABAB.​


First, find the gradient of the line segment ABABAB, using m=y2−y1x2−x1m= \dfrac{y_2-y_1}{x_2 -x_1}m=x2​−x1​y2​−y1​​.

m=3−12−6−−3=3m= \dfrac{3-12}{-6 --3}=3m=−6−−33−12​=3​

So the gradient of its perpendicular bisector is −13-\dfrac{1}{3}−31​.


Next, find the midpoint of AB using ((x1+x2)2,(y1+y2)2){\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}(2(x1​+x2​)​,2(y1​+y2​)​).

((−6+−3)2,(3+12)2)(−92,152){\bigg(\dfrac {(-6 + -3)}{2} , \dfrac {(3 + 12)}{2}\bigg)}\\{\bigg(\dfrac {-9}{2} , \dfrac {15}{2}\bigg)}(2(−6+−3)​,2(3+12)​)(2−9​,215​)

​

Substitute the coordinates of the midpoint ofABABAB​ into y−y1=−13(x−x1)\\\\y-y_1=-\dfrac{1}{3}(x-x_1)y−y1​=−31​(x−x1​) to find the equation of ABABAB.

y−152=−13(x−−92)y−152=−13x−32   y=−13x+6‾\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}y−215​y−215​​=−31​(x−−29​)=−31​x−23​​   y=−31​x+6​


​​​

b) Find the equation of the perpendicular bisector BCBCBC.


First, find the gradient of the line segment BCBCBC​, using m=y2−y1x2−x1m= \dfrac{y_2-y_1}{x_2 -x_1}m=x2​−x1​y2​−y1​​.

m=12−3−3−0=−3m= \dfrac{12-3}{-3 -0}=-3m=−3−012−3​=−3​​

​​

So the gradient of its perpendicular bisector is 13\dfrac{1}{3}31​.

Next, find the midpoint of BCBCBC using ((x1+x2)2,(y1+y2)2){\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}(2(x1​+x2​)​,2(y1​+y2​)​).

((−3+0)2,(12+3)2)(−32,152){\bigg(\dfrac {(-3 + 0)}{2} , \dfrac {(12 + 3)}{2}\bigg)}\\{\bigg(\dfrac {-3 }{2} , \dfrac {15}{2}\bigg)}(2(−3+0)​,2(12+3)​)(2−3​,215​)

​​​

Substitute the coordinates of the midpoint of BCBCBC into y−y1=13(x−x1)\\y-y_1=\dfrac{1}{3}(x-x_1)y−y1​=31​(x−x1​) to find the equation of BCBCBC.

y−152=13(x−−32)y−152=13x+12   y=13x+8‾\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}y−215​y−215​​=31​(x−−23​)=31​x+21​​   y=31​x+8​

​​​

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


Find the point of intersection between perpendicular bisectors ABABAB and BCBCBC.


​Make xxx the subject of BCBCBC:

x=3y−24x=3y-24x=3y−24

Substitute the equation for BCBCBC into the equation for ABABAB and solve for yyy.

y=−13(3y−24)+63y=−(3y−24)+183y=−3y+24+186y=42y=7\begin{aligned}y&=-\dfrac{1}{3}(3y-24)+6 \\3y&=-(3y-24)+18\\3y&=-3y+24+18\\6y&=42\\y&=7\end{aligned}y3y3y6yy​=−31​(3y−24)+6=−(3y−24)+18=−3y+24+18=42=7​​​

​​​​​​

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

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


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


d) Find the equation of the circumcircle. 


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

 d=(x1−x2)2+(y1−y2)2=(−3−−3)2+(12−7)2=0+52=5\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}d​=(x1​−x2​)2+(y1​−y2​)2​=(−3−−3)2+(12−7)2​=0+52​=5​,


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

​

Maths; Circles; KS5 Year 12; Circles and triangles


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FAQs - Frequently Asked Questions

What is a circumcircle?

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

How can the centre of a circle be found using the perpendicular bisectors of two chords?

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. The centre of a circumcircle is the circumcentre.

For a right-angled triangle which side is the diameter of the circle?

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

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