Intersection of straight lines and circles

In a nutshell

Algebra is used to find the coordinates of the point of intersection of a circle and a straight line. Straight lines can intersect a circle once, twice, or not at all. A tangent is a line that touches a circle at one point.

Finding points of intersection 

To find the points of intersection, the equations of the circle and the straight line can be solved simultaneously. 

Example 1

Find the coordinates of the points where the line $$y=x+4$$​ intersects the circle $$\left(x-3\right)^{2}+\left(y-5\right)^{2}\ =\ 34$$.

First, substitute the line equation into the quadratic to solve for $$x$$ or $$y$$.

$$\begin{aligned}\left(x-3\right)^{2}+\left((x+4)-5\right)^{2} &=\ 34 \\\left(x-3\right)^{2}+\left(x-1\right)^{2} &= 34\end{aligned}$$

Expand the brackets to collect terms.

$$\begin{aligned}({x^2} - 6x+9) + (x^2 -2x +1)&=34 \\2x^2-8x+10&=34 \\2x^2-8x-24 &=0 \\x^2 - 4x-12&=0\\(x-6)(x+2)&=0\end{aligned}$$

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Solve for $$x$$.​

$$(x-6)(x+2)=0$$​​

$$x=6$$​ and $$x=-2$$

Substitute the $$x$$ terms into the line equation to solve for $$y$$​. ​

$$y=6+4=10$$

$$y = -2+4 =2$$

Therefore, the line $$y=x+4$$ intersects the circle at points​

$$\underline{(6,10)}$$ and $$\underline{(-2,2)}$$

Finding the number of intersection points

Solve the line and circle equations simultaneously to find the discriminant $$b^2-4ac$$ to test for roots of the quadratic equation. 

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​Example 2

Show that the line $$y= x-7$$ does not meet the circle $$(x+3)^2+y^2=42$$​.

First, substitute line $$y= x-7$$ into the equation of the circle and collect terms.  ​

$$\begin{aligned}(x+3)^2+(x-7)^2&=42 \\x^2+6x+9+x^2-14x+49&=42 \\2x^2-8x+16&=0 \\x^2-4x+8 &= 0\end{aligned}$$​​

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Use the discriminant of the quadratic to test for roots. 

$$b^2-4ac = (-4)^2-(4\times1\times8) = -16$$

 $$-16<0$$, therefore, there are no points of intersection. 

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