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.
Which of the following is the graph of a line and a circle with two points of intersection?
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.
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.
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).
Consider the system of equations.
$x^2+y^2=10$x2+y2=10
$x-y=2$x−y=2
$\left(x,y\right)$(x,y) is a solution to the system of equations. First solve for $x$x.
Therefore, the solutions are $($($3$3, $\editable{}$$)$) and $($($-1$−1, $\editable{}$$)$).
Consider the equations $x^2+y^2=100$x2+y2=100 and $y=x-2$y=x−2.
What kind of graph is represented by the equation $x^2+y^2=100$x2+y2=100?
a straight line
a parabola
a cubic
a circle
Determine the radius of the circle.
Determine the coordinates of the centre of the circle.
On the same set of axes, graph $x^2+y^2=100$x2+y2=100 and $y=x-2$y=x−2.
State the coordinates of the points of intersection.