EXT: Systems of straight lines and circles

In this section, we consider systems of two equations where one of them is linear and the other describes a circle.

Linear equations in two unknowns can be written in the form $ax+y=c$ax+y=c. (This can be rearranged to the possibly more familiar form, $y=mx+c$y=mx+c if desired.)

Circles have the general form $(x-a)^2+(y-b)^2=r^2$(x−a)2+(y−b)2=r2 where the centre of the circle is the point $(a,b)$(a,b) and $r$r is the radius. (This equation is derived by considering a circle in the Cartesian plane and applying the Pythagoras theorem, you can read about it here.)

Have a play with this applet.  It shows a straight line and a circle.  Move the straight line around and consider these questions.

1) Is there a situation where a straight line may not intersect a circle?  How is this different to when it intersects another linear line?

2) Is there a situation where a straight line may intersect only once?

3) Is there a situation where a straight line may intersect more than once, if so what is the maximum number of times?

 

As we have just seen, a straight line may or may not intersect a given circle. This means that a system of equations involving a line and a circle may or may not have a solution.

In fact, there may be

zero solutions (where the line doesn't cross)
one solution (where it just touches)
two solutions (where it passes through the circle)

As with systems with a line and a hyperbola, we will find that a quadratic equation is involved in solving a system with a line and a circle. The discriminant of the quadratic determines the existence and number of the solutions. 

Worked Example 1

 

Some further examples

Question 1

Consider the system of two equations

$y-x$y−x	$=$=	$1$1
$y^2+x^2$y2+x2	$=$=	$3$3

This can be solved by writing the first equation in the form $y=1+x$y=1+x and substituting this expression for $y$y into the second equation. This gives $(1+x)^2+x^2=3$(1+x)2+x2=3 which is a quadratic equation in $x$x. So, after expanding the bracket and collecting like-terms, we have $2x^2+2x-2=0$2x2+2x−2=0 or $x^2+x-1=0$x2+x−1=0. By the quadratic formula, the solutions for $x$x are 

$x=\frac{-1\pm\sqrt{5}}{2}$x=−1±√52​

When these values for $x$x are substituted into the first equation we find that the corresponding values for $y$y are

$y=\frac{1\pm\sqrt{5}}{2}$y=1±√52​

We should now check that these $(x,y)$(x,y) points do satisfy the equations in the system.

 

question 2

Find lines $y=2x+b$y=2x+b that are tangent to the circle $x^2+(y-1)^2=4$x2+(y−1)2=4.

In other words, we have to find one or more values of $b$b that fulfil this requirement.

We substitute $2x+b$2x+b for $y$y in the circle equation. Thus,

$x^2+(2x+b-1)^2$x2+(2x+b−1)2	$=$=	$4$4
$x^2+(2x+b)^2-2(2x+b)+1$x2+(2x+b)2−2(2x+b)+1	$=$=	$4$4
$x^2+4x^2+4bx+b^2-4x-2b+1$x2+4x2+4bx+b2−4x−2b+1	$=$=	$4$4
$5x^2+4x(b-1)+(b-1)^2-4$5x2+4x(b−1)+(b−1)2−4	$=$=	$0$0

This is a quadratic in $x$x with $b$b yet to be determined. According to the quadratic formula,

$x=\frac{-4(b-1)\pm\sqrt{16(b-1)^2-20\left((b-1)^2-4\right)}}{10}$x=−4(b−1)±√16(b−1)2−20((b−1)2−4)10​

The expression under the square root sign is the discriminant. If it has the value zero then there is a single solution for $x$x, which means the line is tangent to the circle. We look for the values of $b$b that make this happen.

We want $b$b such that $16(b-1)^2-20\left((b-1)^2-4\right)=0$16(b−1)2−20((b−1)2−4)=0. Thus,

$-4(b-1)^2+80$−4(b−1)2+80	$=$=	$0$0
$(b-1)^2-20$(b−1)2−20	$=$=	$0$0
$(b-1+\sqrt{20})(b-1-\sqrt{20})$(b−1+√20)(b−1−√20)	$=$=	$0$0

Thus, the solutions for $b$b are $b=1-2\sqrt{5}$b=1−2√5 and $b=1+2\sqrt{5}$b=1+2√5.

This means the lines $y=2x+1-2\sqrt{5}$y=2x+1−2√5  and $y=2x+1+2\sqrt{5}$y=2x+1+2√5 should both be tangent to the circle $x^2+(y-1)^2=4$x2+(y−1)2=4. The graphs are shown below.

 

question 3

What are the points of tangency in Example 2?

We found the solutions $x=\frac{-4(b-1)\pm\sqrt{16(b-1)^2-20\left((b-1)^2-4\right)}}{10}$x=−4(b−1)±√16(b−1)2−20((b−1)2−4)10​ and we determined that we could have $b=1-2\sqrt{5}$b=1−2√5 or $b=1+2\sqrt{5}$b=1+2√5 as these values of $b$b would make the disctiminant zero.

Thus, the solutions for $x$x must be 

$x=\frac{-4\times(-2\sqrt{5})}{10}$x=−4×(−2√5)10​ for $b=1-2\sqrt{5}$b=1−2√5 and $x=\frac{-4\times2\sqrt{5}}{10}$x=−4×2√510​ for $b=1+2\sqrt{5}$b=1+2√5. That is, 

$x=\frac{4\sqrt{5}}{5}$x=4√55​  and  $x=-\frac{4\sqrt{5}}{5}$x=−4√55​.

The corresponding values for $y$y are 

$y$y	$=$=	$2\times\frac{4\sqrt{5}}{5}+1-2\sqrt{5}$2×4√55​+1−2√5
	$=$=	$1-\frac{2\sqrt{5}}{5}$1−2√55​

and

$y$y	$=$=	$2\times\left(-\frac{4\sqrt{5}}{5}\right)+1+2\sqrt{5}$2×(−4√55​)+1+2√5
	$=$=	$1+\frac{2\sqrt{5}}{5}$1+2√55​

 

As decimals, the points of tangency are approximately $(1.79,0.11)$(1.79,0.11)  and  $(-1.79,1.89)$(−1.79,1.89).

question 4

Question 5