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 $$d=\sqrt{{(x_{1} - x_{2}})^2+{(y_{1} - y_{2})^2}}$$.

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

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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 $$\underline{y=\dfrac{8}{5}x+\dfrac{13}{5} }$$

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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 $$C$$.

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

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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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