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Using tangent and chord properties

Using tangent and chord properties

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Summary

Using tangent and chord properties

In a nutshell

The properties of tangents and chords are very useful when solving geometric problems. At the point of intersection, a tangent to a circle is perpendicular to the radius of the circle. A chord is a line segment joining two points on a circle. The perpendicular bisector of a chord passes through the centre of a circle. 



Using tangent properties to solve problems 

A tangent is a straight line that intersects a circle at one point, and is perpendicular to the radius of the circle at the point of intersection.


Example 1

The circle CCC has centre M(12,4)M(12,4)M(12,4) and passes through point P(4,9)P(4,9) P(4,9).


a) Find the equation for CCC​. 


Using the equation, (x−a)2+(y−b)2=r2{{(x-a)^2} + {(y-b)^2} = {r^2}}(x−a)2+(y−b)2=r2, where (a,b)(a,b)(a,b) are the coordinates of the centre of the circle and rrr is the radius, the circle has the equation (x−12)2+(y−4)2=r2{{(x-12)^2} + {(y-4)^2} = {r^2}}(x−12)2+(y−4)2=r2.


To find the radius of the circle, the distance between the centre MMM​ and the point PPP must be found. 


To find the distance between two points use the equation d=(x1−x2)2+(y1−y2)2d=\sqrt{{(x_{1} - x_{2}})^2+{(y_{1} - y_{2})^2}}d=(x1​−x2​)2+(y1​−y2​)2​.

d=(12−4)2+(4−9)2=(8)2+(−5)2=64+25=89\begin{aligned}d&=\sqrt{{(12 - 4})^2+{(4 - 9)^2}}\\ &=\sqrt{{(8})^2+{(-5)^2}} \\ &=\sqrt{64+25}\\ &=\sqrt{89}\end{aligned}d​=(12−4)2+(4−9)2​=(8)2+(−5)2​=64+25​=89​​

​​​

As the radius is 89\sqrt{89}89​, the equation of CCC is

(x−12)2+(y−4)2=89‾\underline{{(x-12)^2} + {(y-4)^2} = 89}(x−12)2+(y−4)2=89​


The line lll is a tangent to CCC​ and passes through point PPP. 


b) Find the equation for lll.​


The radius between point PPP and the centre of the circle MMM is perpendicular to the tangent line lll.


To find equation of line lll, first find the gradient between points PPP and MMM using m=y1−y2x1−x2m= \dfrac{y_1-y_2}{x_1 -x_2}m=x1​−x2​y1​−y2​​.

m=9−44−12=−58m= \dfrac{9-4}{4 -12} = -\dfrac{5}{8} m=4−129−4​=−85​


So line lll has a gradient of 85\dfrac{8}{5} 58​ , as the product of perpendicular gradients is −1-1−1.


Now, find the equation of line lll by substituting in the point (4,9)(4,9)(4,9).​​

y−y1=m(x−x1)y−9=85(x−4)y−9=85x−325\begin{aligned}y-y_1&=m(x-x_1)\\\\y-9&=\dfrac{8}{5} (x-4)\\\\y-9&=\dfrac{8}{5}x - \dfrac{32}{5} \end{aligned}y−y1​y−9y−9​=m(x−x1​)=58​(x−4)=58​x−532​​


The equation of line lll is y=85x+135‾\underline{y=\dfrac{8}{5}x+\dfrac{13}{5} }y=58​x+513​​

Maths; Circles; KS5 Year 12; Using tangent and chord properties

​


Using chord properties to solve problems 

Chords are line segments that join two points on the circumference of a circle. The perpendicular bisector of a chord passes through the centre of a circle.


Example 2

The points A(4,−2)A(4,-2)A(4,−2) and B(13,−5)B(13,-5)B(13,−5) lie on a circle with centre CCC.MMM is the midpoint of the line segment ABABAB. The line lll passes through both points MMM and CCC.

a) Find the line equation for lll.


First find the gradient of the line segment ABABAB using m=y1−y2x1−x2m= \dfrac{y_1-y_2}{x_1 -x_2}m=x1​−x2​y1​−y2​​.

m=−5−−213−4=−39=−13m= \dfrac{-5--2}{13 -4}= -\dfrac{3}{9} =-\dfrac{1}{3}m=13−4−5−−2​=−93​=−31​


The gradient of the line lll must be 333, as the product of perpendicular gradients is −1.-1. −1.


Find the coordinates of the point MMM by finding the midpoint of ABABAB using the equation ((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​)​).

((4+13)2,(−2+−5)2)=(172,−72)=(8.5,−3.5){\bigg(\dfrac {(4 + 13)}{2} , \dfrac {(-2 + -5)}{2}\bigg)} = {\bigg(\dfrac {17}{2} , \dfrac {-7}{2}\bigg)}=(8.5, -3.5)(2(4+13)​,2(−2+−5)​)=(217​,2−7​)=(8.5,−3.5)​


Find the equation of line lll​ by substituting the gradient of 333 and the coordinates of point MMM​ into y−y1=m(x−x1)y-y_1=m(x-x_1)y−y1​=m(x−x1​)​.

y−−3.5=3(x−8.5)y+3.5=3x−512y--3.5=3(x-8.5)\\y+3.5=3x-\dfrac{51}{2}y−−3.5=3(x−8.5)y+3.5=3x−251​


Therefore, the equation of line lll is y=3x−29‾\underline{y=3x-29}y=3x−29​.


b) Given that the xxx-coordinate of CCC is 888, find the yyy-coordinate of CCC.


Line lll passes through CCC, so substitute x=8x=8x=8 into y=3x−29{y=3x-29}y=3x−29 to find the yyy-coordinate of CCC.

y=3(8)−29y=−5‾\begin{aligned}y&=3(8)-29\\y&=\underline{-5}\end{aligned}yy​=3(8)−29=−5​​

​​​

c) Find the equation of the circle. 


As the centre CCC has coordinates (8,−5)(8,-5)(8,−5), the equation of the circle will be (x−8)2+(y+5)2=r2{{(x-8)^2} + {(y+5)^2} = r^2}(x−8)2+(y+5)2=r2.


To find the radius of the circle, the distance between point AA A and CCC can be found using d=(x1−x2)2+(y1−y2)2d=\sqrt{{(x_{1} - x_{2}})^2+{(y_{1} - y_{2})^2}}d=(x1​−x2​)2+(y1​−y2​)2​.

d=(8−4)2+(−5−−2)2=(4)2+(−3)2=16+9=25=5\begin{aligned}d&=\sqrt{{(8 - 4})^2+{(-5 --2)^2}}\\&=\sqrt{{(4})^2+{(-3)^2}}\\&=\sqrt{16+9}\\&=\sqrt{25}\\&=5\end{aligned}d​=(8−4)2+(−5−−2)2​=(4)2+(−3)2​=16+9​=25​=5​


​Therefore the equation of the circle is (x−8)2+(y+5)2=25‾\underline{{(x-8)^2} + {(y+5)^2} = 25}(x−8)2+(y+5)2=25​.​

​

Maths; Circles; KS5 Year 12; Using tangent and chord properties
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FAQs - Frequently Asked Questions

What does the perpendicular bisector of a chord pass through?

The perpendicular bisector of a chord passes through the centre of a circle.

What is a chord?

A chord is a line segment joining two points on a circle.

What is the tangent of a circle perpendicular to?

At the point of intersection a tangent to a circle is perpendicular to the radius of the circle.

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