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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 $C$ has centre $M(12,4)$ and passes through point $P(4,9)$ .

a) Find the equation for $C$ .

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

To find the radius of the circle, the distance between the centre $M$  and the point $P$ must be found.

To find the distance between two points use the equation $d=\sqrt{{(x_{1} - x_{2}})^2+{(y_{1} - y_{2})^2}}$ .

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

As the radius is $\sqrt{89}$ , the equation of $C$ is

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

The line $l$ is a tangent to $C$  and passes through point $P$ .

b) Find the equation for $l$ .

The radius between point $P$ and the centre of the circle $M$ is perpendicular to the tangent line $l$ .

To find equation of line $l$ , first find the gradient between points $P$ and $M$ using $m= \dfrac{y_1-y_2}{x_1 -x_2}$ .

$m= \dfrac{9-4}{4 -12} = -\dfrac{5}{8}$

So line $l$ has a gradient of $\dfrac{8}{5}$ , as the product of perpendicular gradients is $-1$ .

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

$\begin{aligned}y-y_1&=m(x-x_1)\\\\y-9&=\dfrac{8}{5} (x-4)\\\\y-9&=\dfrac{8}{5}x - \dfrac{32}{5} \end{aligned}$

The equation of line $l$ is $\underline{y=\dfrac{8}{5}x+\dfrac{13}{5} }$

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 $A(4,-2)$ and $B(13,-5)$ lie on a circle with centre $C$ . $M$ is the midpoint of the line segment $AB$ . The line $l$ passes through both points $M$ and $C$ .

a) Find the line equation for $l$ .

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

$m= \dfrac{-5--2}{13 -4}= -\dfrac{3}{9} =-\dfrac{1}{3}$

The gradient of the line $l$ must be $3$ , as the product of perpendicular gradients is $-1.$

Find the coordinates of the point $M$ by finding the midpoint of $AB$ using the equation ${\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}$ .

${\bigg(\dfrac {(4 + 13)}{2} , \dfrac {(-2 + -5)}{2}\bigg)} = {\bigg(\dfrac {17}{2} , \dfrac {-7}{2}\bigg)}=(8.5, -3.5)$

Find the equation of line $l$ by substituting the gradient of $3$ and the coordinates of point $M$ into $y-y_1=m(x-x_1)$ .

$y--3.5=3(x-8.5)\\y+3.5=3x-\dfrac{51}{2}$

Therefore, the equation of line $l$ is $\underline{y=3x-29}$ .

b) Given that the $x$ -coordinate of $C$ is $8$ , find the $y$ -coordinate of $C$ .

Line $l$ passes through $C$ , so substitute $x=8$ into ${y=3x-29}$ to find the $y$ -coordinate of $C$ .

$\begin{aligned}y&=3(8)-29\\y&=\underline{-5}\end{aligned}$

c) Find the equation of the circle.

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

To find the radius of the circle, the distance between point $A$ and $C$ can be found using $d=\sqrt{{(x_{1} - x_{2}})^2+{(y_{1} - y_{2})^2}}$ .

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

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

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