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The mutual position of a circle and a straight line; the common points of two circles

The mutual position of a circle and a straight line

A circle and a straight line can be located in the plane in three mutual
positions (Figure 47):

they have no point in common;
they have one point in common (the straight line touches the circle);
they have two points in common.

\latex{e\cap k=\varnothing}

\latex{e\cap k=E}

\latex{e\cap k=(M_{1};M_{2} )}

\latex{ k }

\latex{ e }

\latex{ E }

\latex{ M_2 }

\latex{ e }

\latex{ M_1 }

Figure 47

Example 1

Let us give the straight lines passing through the origin with which the circle with the equation \latex{(x+2)^{2}+(y-3)^{2}=3} has

\latex{0}

\latex{1}

\latex{2} points(s) common.

Solution

The equation of the straight lines passing through the origin, except for the \latex{ y }-axis, has the form of \latex{y=mx} (Since the equation of the \latex{ y }-axis is \latex{x=0}, and in this case the expression on the left-hand-side of the equation of the circle is not less than \latex{ 4 }, the \latex{ y }-axis does not have a point which lies on the circle (Figure 48))

The coordinates of the common points of such a straight line and the circle satisfy both the equation of the circle and the equation of the straight line, thus the example translated to the language of algebra is as follows:

For which real values of the parameter m will the quadratic parametric
simultaneous equations

\latex{\begin{rcases}(x-2)^{2}+(y-3)^{2}=3 \\ y=mx\end{rcases};}

(1)

have \latex{a)0\,;\,b)\,1;\,c)\,2} real number pairs of solution?

Let us substitute the second equation into the first one:

\latex{(x-2)^{2}+(mx-3)^{2}=3}

By squaring and rearranging it so that one side is equal to \latex{ 0 }:

\latex{(m^{2}+1 )x^{2}-2\times (3m\times 2)x+10=0} (2)

This parametric quadratic equation has \latex{ 0 }, \latex{ 1 } or \latex{ 2 } real solution(s) depending on the discriminant being negative, \latex{ 0 } or positive. So let us examine the discriminant of the equation.

\latex{D=\left[-2\times (3m+2) \right]^{2}-4\times (m^{2}+1 )\times 10=4\times \left[(3m+2)^{2}-10\times (m^{2}+1 ) \right].}

The sign of \latex{ D } is determined by the quadratic expression in terms of \latex{ m } inside the square brackets. When rearranging it:

\latex{(3m+2)^{2}-10\times (m^{2}+1 )=9m^{2}+12m+4-10m^{2}-10=-m^{2}+12m-6.}

Let us first determine when the discriminant will be \latex{ 0 }. For this we have to solve the quadratic equation

\latex{-m^{2}+12m-6=0.}

After substituting into the formula, we get

\latex{m_{1}=6+\sqrt{30}\approx 11.48} and \latex{m_{2}=6-\sqrt{30}\approx 0.52.}

In these two cases the simultaneous equations above have one ordered real number pair as solution, which geometrically means that the straight lines with the equations

\latex{y=(6+\sqrt{30} )\times x} and \latex{y=(6-\sqrt{30} )\times x}

are the tangent lines of the circle passing through the origin. With this we answered part b) of the example.

The circle and the straight line do not have a point in common exactly when the simultaneous equations (1) have no real solutions. However it holds exactly when the discriminant of the parametric equation (2) obtained for \latex{ x } is negative. It is equivalent to the following:

\latex{-m^{2}+12m-6\lt 0.}

As the leading coefficient of the quadratic expression in terms of \latex{ m } on the left-hand-side is negative, the value of the expression is negative exactly when

\latex{m\lt 6-\sqrt{30}} or \latex{6+\sqrt{30}\lt m} (Figure 49). (3)

So the straight lines passing through the origin the gradients of which satisfy one of the inequalities (3), do not have a point in common with the circle (Figure 50). (We have already referred to that the \latex{ y }-axis does not have a point in common with the circle either.) This is the answer to part a) of the example.

The straight line intersects the circle at two points exactly when the simultaneous equations (1) have two distinct ordered real number pairs as solution. It is satisfied exactly when the discriminant of the equation (2) is positive, i.e.

\latex{-m^{2}+12m-6\gt 0.}

In Figure 49 it can be seen that it is satisfied if and only if

\latex{6-\sqrt{30}\lt m\lt 6+\sqrt{30}.} (4)

So the straight lines passing through the origin whose gradients satisfy the inequality (4) have two points in common with the circle. With this we also answered question c).

Generalisation

The coordinates of the common points of the circle with the equation \latex{(x-u)^{2}+(y-v)^{2}=r^{2}} and the straight line \latex{y=mx+b} are the solutions of the simultaneous equations

\latex{\begin{rcases}(x-u)^{2}+(y-v)^{2}=r^{2} \\ y=mx+b\end{rcases}.}

The discriminant of the quadratic equation in one variable obtained after substituting the second equation into the first one (eliminating one of the unknowns) determines the number of common points:

if the discriminant is positive, then the straight line intersects the circle at two points;
if the discriminant is \latex{ 0 }, the straight line touches the circle;
if the discriminant is negative, then the straight line and the circle have no points in common.

Example 2

Let us calculate the radius of the circle with the centre \latex{K(-3;4),} which touches the straight line with the equation \latex{x+2y=-5} et us give the coordinates of the tangent point.

Solution I

The circle in question and the straight line given have one point in common (Figure 51), which algebraically means that the simultaneous equations

\latex{\begin{rcases}(x+3)^{2}+(y-4)^{2}=r^{2} \\ x+2y=-5\end{rcases}}

have one real number pair (the coordinates of the tangent point) as solution.
From the second equation \latex{x=-2y-5.} By substituting it into the first equation

\latex{(-2y-2)^{2}+(y-4)^{2}=r^{2},}

and after rearranging

\latex{5y^{2}+20-r^{2}=0.}

This incomplete quadratic equation has one solution exactly when
\latex{r^{2}=20,} i.e. \latex{r=\sqrt{20}=2\times \sqrt{5}\approx 4.47.} Then

\latex{y=0} and \latex{x=-5}

The radius in question is: \latex{r=2\times \sqrt{5};} the tangent point is: \latex{E(-5;0)}

Solution II

During the previous solution we had an algebraic approach to the problem; now we are working based on geometric considerations.

The radius passing through the tangent point \latex{ E } is perpendicular to the tangent line, so the equation of the straight line \latex{ KE } can be given in view of the equation of the tangent line. \latex{\overrightarrow{n}(1;2)} is a normal vector of the tangent line. It is a direction vector of the straight line \latex{ KE }, thus the equation of \latex{ KE } can be set up:

\latex{\begin{rcases}\overrightarrow{v}(1;2) \\ K(-3;4) \end{rcases} 2x-y=-10}

\latex{ E } is the intersection point of the tangent line and the straight line \latex{ KE }, thus its coordinates are obtained as the solution of the simultaneous equations

\latex{\begin{rcases}x+2y=-5 \\ 2x-y=-10\end{rcases}}.

By solving the simultaneous equations the tangent point is obtained:
\latex{E(-5;0)}. Thus the radius of the circle is:

\latex{r=\sqrt{(-3-(-5))^{2}+(4-0)^{2} }=\sqrt{20}=2\times \sqrt{5}.}

The intersection points of two circles

The number of the common points of two circles (boundary lines of two circles) in the plane can be \latex{ 0 }, \latex{ 1 } or \latex{ 2 } (Figure 52). Similarly to the case of two straight lines or a circle and a straight line the number of the common points algebraically again depends on the number of real number pairs the simultaneous equations consisting of the equations of the two circles have as solution.

Example 3

Let us calculate the coordinates of the intersection points of the circle \latex{k_{1}} with the equation \latex{(x-2)^{2}+y^{2}=20} and the circle \latex{k_{2}} with the equation \latex{(x+7)^{2}+(y+3)^{2}=50}.

Solution

The coordinates of the common points of the two circles satisfy both
equations; therefore these are solutions of the simultaneous equations
below:

\latex{\begin{rcases}(x-2)^{2}+y^{2}=20 \\ (x+7)^{2}+(y+3)^{2}=50 \end{rcases}}.

For the solution it is practical to perform the operations of squaring in the equations and then to rearrange the equations so that one side is equal to \latex{ 0 }:

\latex{\begin{rcases}x^{2}+y^{2}-4x-16=0 \\ x^{2}+y^{2}+14x+6y+8=0\end{rcases}.}

If we subtract one equation from the other one, then the quadratic terms will be removed, thus we can express one unknown in terms of the other one. According to this let us now subtract the first equation from the second one.

\latex{18x+6y+24=0}

By dividing both sides by \latex{ 6 }

\latex{3x+y+4=0}

It implies that \latex{y=-3x-4.} Let us substitute it into the first equation.

\latex{x^{2}+(-3x-4)^{2}-4x-16=0}

After performing the squaring and rearranging the following is obtained

\latex{10x^{2}+20x=0.}

By dividing both sides by 10, then by factoring out \latex{ x }

\latex{x(x+2)=0,}

which implies

\latex{x_{1}=-2,x_{2}=0.}

By substituting these into the equation expressing \latex{ y } we obtain the corresponding second coordinates:

\latex{y_{1}=-2,y_{2}=-4.}

So the intersection points of the two circles are \latex{M_{1}(-2;2),M_{2}(0;-4)} (Figure 53).

\latex{y=-3x-4}

\latex{K_{1}(2;0)}

\latex{K_{2}(-7;-3)}

\latex{M_{1}(-2;2)}

\latex{M_{2}(0;-4)}

\latex{ x }

\latex{ K_{2} }

\latex{ 1 }

\latex{ 0 }

\latex{ 1 }

\latex{ y }

\latex{ K_{1} }

Figure 53

Note: During the solution of example \latex{ 3 }, when we subtracted the equations of the two circles from each other, we got the linear equation in two variables \latex{3x+y+4=0} which is the equation of a straight line. As the coordinates of the intersection points of the two circles satisfy the equations of both circles, these also satisfy the equation obtained by subtracting the two equations. And, as two points unambiguously define a straight line, it means that the equation \latex{3x+y+4=0} is the equation of the common secant line of the two circles.

Example 4

How long is the radius of the circle \latex{k_{1}} with the centre \latex{K_{1}(-4;1),} if it touches the circle \latex{k_{2}} with the equation \latex{(x-2)^{2}+(y-9)^{2}=4?} Let us calculate the coordinates of the tangent point.

Solution

Two circles satisfy the conditions of the example, since \latex{k_{2}} can touch \latex{k_{1}} from outside and from inside too (Figure 54).

\latex{K_{2}(2;9)}

\latex{K_{1}(-4;1)}

\latex{ x }

\latex{ 1 }

\latex{ 0 }

\latex{ 1 }

\latex{ y }

\latex{ K_{1} }

\latex{ r_{2} }

\latex{ K_{2} }

\latex{ E }

\latex{ K_{1} }

\latex{ r_{1} }

Figure 54

Let us first consider the case when it touches it from outside. If \latex{r_{1}} denotes the radius of the circle \latex{k_{1}}, \latex{r_{2}} denotes the radius of the circle \latex{k_{2}}, and \latex{K_{2}} denotes the centre of the circle \latex{k_{2}}, then

\latex{r_{1}+r_{2}=K_{1}K_{2}.}

The coordinates of \latex{K_{2}} and \latex{r_{2}} can be given from the equation of the circle \latex{k_{2}:K_{2}(2;9),r_{2}=2.}

\latex{r_{1}=K_{1}K_{2}-r_{2}=\sqrt{(2-(-4)^{2}+(9-1)^{2} )}-2=10-2=8.}

So in this case \latex{r_{1}=8} thus the equation of \latex{k_{1}} is

\latex{(x+4)^{2}+(y-1)^{2}=64.}

To determine the coordinates of the tangent points we have to solve the simultaneous equations

\latex{\begin{rcases}(x-2)^{2}+(y-9)^{2}=4 \\ (x+4)^{2}+(y-1)^{2}=64\end{rcases}}.

By performing the operations of squaring and then by rearranging the equation so that one side is equal to

\latex{\begin{rcases}x^{2}+y^{2}-4x-18y+81=0 \\ x_{2}+y_{2}+8x-2y-47=0\end{rcases}}.

By subtracting the first equation from the second one

\latex{12x+16y-128=0,}

then by dividing both sides by \latex{ 4 }

\latex{3x+4y-32=0.}

This implies that \latex{x=\frac{32-4y}{3}}. Let us substitute it into the first equation:

\latex{(\frac{32-4y}{3} )^{2}+y^{2}-4\times \frac{32-4y}{3}-18y+81=0.}

By performing the operations and by rearranging we obtain

\latex{25y^{2}-370y+1369=0.}

There is \latex{(5y-37)^{2}} on the left-hand-side, and it is equal to \latex{ 0 } if and only if \latex{y=\frac{37}{5}=7.4.} It implies that \latex{=\frac{32-4\times 7.4}{3}=0.8.} We get that if the two circles touch each other from outside, then the tangent point is \latex{E(0.8;7.4).}

Let \latex{k^{'}_{1}} denote the circle with the centre \latex{K_{1}} and with the radius \latex{r^{'}_{1}} which is touched \latex{k_{2}} from inside.

It can easily be seen that \latex{r^{'}_{1}=r_{1}+2r_{2}=12.}

The equation of the circle \latex{k_{1}} is:

\latex{(x+4)^{2}+(y-1)^{2}=144.}

Now we are determining the coordinates of the tangent point \latex{E^{'}(x;y)} by paying attention to the fact that \latex{K_{2}} bisects the line segment \latex{EE^{'}} Thus

\latex{2=\frac{0.8+x}{2}} and \latex{9=\frac{7.4+y}{2},}

which implies \latex{x=3.2} and \latex{y=10.6.} So the tangent point of \latex{k^{'}_{1}} and \latex{k_{2}} is \latex{E^{'}(3.2;10.6).}

Note: In the first part of the solution we also could have determined the
coordinates of \latex{ E } based on the relation relating to the coordinates of the point of division. Now, as a check, we calculate them this way too.

E divides the line segment \latex{K_{1}K_{2}} into two in the ratio of \latex{r_{1}:r_{2},} i.e.

\latex{\frac{K_{1}E }{EK_{2} }=\frac{r_{1} }{r_{2} }=\frac{8}{2}=4.}

Thus the first coordinate of \latex{ E } is

\latex{\frac{-4+4\times 2}{5}=\frac{4}{5}=0.8,}

its second coordinate is

\latex{\frac{1+4\times 9}{5}=\frac{37}{5}=7.4.}

Now we are giving the tangent lines that can be drawn to a given circle from a given point outside the circle with the tools of coordinate geometry in connection with an example. We solved this problem in year \latex{ 9 } with a pair of compasses and a ruler, i.e. we showed how to construct the tangent lines.

Example 5

Let us give the equation of the tangent lines that can be drawn to the circle with the equation \latex{(x-5)^{2}+(y-1)^{2}=10} from the point \latex{P(-2;2).}

Solution I

In this solution we are following the same path we followed in the corresponding construction exercise in year \latex{ 9 }.

The centre of the circle is \latex{K(5;1)} , its radius is \latex{r=\sqrt{10}} . Since the radius drawn to the tangent point is perpendicular to the tangent line, Thales' theorem implies that the circle drawn with the line segment \latex{ KP } as the diameter intersects the given circle at the tangent points. The tangent points and the point \latex{ P } unambiguously define the tangent lines
(Figure 55).

\latex{P(-2;2)}

\latex{K(5;1)}

\latex{ e_{1} }

\latex{ y }

\latex{ 1 }

\latex{ 0 }

\latex{ x }

\latex{ F }

\latex{ E_{1} }

\latex{ E_{2} }

\latex{ e_{2} }

\latex{ 1 }

Figure 55

The steps of the solution are:

The midpoint of \latex{ KP } is \latex{F(\frac{3}{2};\frac{3}{2} ).}
\latex{FP=\frac{KP}{2}=\frac{1}{2}\times \sqrt{7^{2}+(-1)^{2} }=\frac{\sqrt{50} }{2}=\frac{5\times \sqrt{2} }{2}.}
The equation of the Thales' circle with the centre \latex{ F } is:

\latex{(x-\frac{3}{2} )^{2}+(y-\frac{3}{2} )^{2}=\frac{25}{2}.}

The simultaneous equations to solve:

\latex{\begin{rcases}(x-5)^{2}+(y-1)^{2}= 10 \\ (x-\frac{3}{2} )^{2}+(y-\frac{3}{2} )^{2}=\frac{25}{2}\end{rcases}.}

By performing the operations of squaring and by rearranging the equation so that one side is equal to

\latex{\begin{rcases}x^{2}+y^{2}-10x-2y+16=0 \\ x^{2}+y^{2}-3x-3y-8=0\end{rcases}.}

Let us subtract the first equation from the second one:

\latex{7x-y-24=0,}

which implies \latex{7x-y-24=0,} By substituting this into the second equation, then by rearranging the equation obtained we get that

\latex{5x^{2}-36x+64=0.}

Its roots are: \latex{x_{1}=4,x_{2}=\frac{16}{5}.} By substituting these into the expression of y the following is obtained: \latex{y_{1}=4,y_{2}=\frac-{8}{5}.} The tangent points are: \latex{E_{1}(4;4),E_{2}(\frac{16}{5};-\frac{8}{5} ).}

Two points of the tangent line \latex{e_{1}} passing through the tangent point \latex{E_{1}} are P and \latex{E_{1}} thus its gradient is

\latex{m_{1}=\frac{4-2}{4-(-2)}=\frac{1}{3}.}

By using the gradient form the equation of \latex{e_{1}} is:

\latex{y-2=\frac{1}{3}\times (x+2),}

or in another form

\latex{x-3y+8=0}

The gradient of the tangent line \latex{e_{2}} passing through the point \latex{E_{2}} is

\latex{m_{2}=\frac{-\frac{8}{5}-2 }{\frac{16}{5}-(-2) }=-\frac{9}{13},}

thus its equation is

\latex{y-2=-\frac{9}{13}\times (x+2)}.

By rearranging it

\latex{9x+13y-8=0}

So the equations of the two tangent lines are

\latex{e_{1}:x-3y+8=0,}
\latex{e_{2}:9x+3y-8=0.}

setting up the equation of the circle with the centre \latex{ F } and with the radius \latex{ KF = FP }

setting up the equations of the straight lines \latex{PE_{1}} and \latex{PE_{2}} calculating the coordinates of the tangent points \latex{E_{1}} and \latex{E_{2}} by solving the simultaneous equations consisting of the equations of the given circle and the Thales’ circle drawn above the line segment \latex{ KP }

Solution II

The method used during the solution of example \latex{ 1 } can also be applied here. Based on the gradient form the equations of the straight lines passing through the point \latex{P(-2;2)} have the form

\latex{y-2=m(x+2),}

where m is the gradient of the straight line. We are looking for the values \latex{ m } for which the circle and the straight line passing through the point \latex{ P } and with the gradient \latex{ m } have exactly one point in common, i.e. the simultaneous equations

\latex{\begin{rcases}(x-5)^{2}+(y-1)^{2}=10 \\ y-2=m\times (x+2)\end{rcases}}

have exactly one real number pair as a solution.

By expressing yfrom the second equation and then by substituting it into the first equation we obtain the parametric quadratic equation

\latex{(m^{2}+1 )x^{2}+2\times (2m^{2}+m-5 )x+4\times (m^{2}+m+4 )=0.}

It has \latex{ 1 } solution exactly when its discriminant is \latex{ 0 }, i.e.

\latex{4\times (2m^{2}+m-5 )^{2}-16\times (m^{2}+1 )\times (m^{2}+m+4 )=0.}

After performing the operations and after rearranging (slightly long calculations) the quadratic equation

\latex{39m^{2}+14m-9=0}

is obtained, the roots of which are

\latex{m_{1} =\frac{1}{3}} and \latex{m_{2} =-\frac{9}{13}}

in accordance with solution \latex{ I }. From here the equation of the tangent lines can be set up as seen before.

Example 6

Let us give the equation of the tangent line passing through the point \latex{P(5;-1)} of the circle with the equation \latex{(x-2)^{2}+(y-3)^{2}=25.}

Solution

The centre of the circle is \latex{K(2;3)} , its radius is \latex{r=5} (Figure 56).
The tangent line e passing through the point \latex{ P } is perpendicular to the straight line \latex{ KP }, therefore the following is a normal vector:

\latex{\overrightarrow{PK}(-3;4).}

So a normal vector and a point (the tangent point) of the tangent line are given, thus its equation is

\latex{-3x+4y=(-3)\times 5+4\times (-1)=-19,}

or in another form

\latex{3x-4y=19.}

Exercises

{{exercise_number}}. Calculate the coordinates of the common points of the circle with the equation \latex{x^{2}+y^{2}=9} and the straight line \latex{ e }, if the equation of \latex{ e } is

\latex{y=x;}

\latex{x+y=0;}

\latex{y=-x+2;}

\latex{y-x=3\times \sqrt{2};}

\latex{3x-y+8=0.}

{{exercise_number}}. Calculate the radius of the circle with the centre \latex{K(3;-2)} if we know that it touches

the \latex{ x }-axis;

the \latex{ y }-axis;

the straight line with the equation \latex{3x– 4y = 6.}

Give the coordinates of the tangent point in each case.

{{exercise_number}}. Give the equation of the circle that touches the coordinate axes and passes through the point \latex{ P }, if

\latex{P(0;1);}

\latex{P(2;3);}

\latex{P(-1;4);}

\latex{P(5;-2);}

\latex{P(-3;-4);}

\latex{P(-2;5)}

{{exercise_number}}. Give the equation of the circle with a radius of \latex{ 4 } units, which touches the straight line with the equation \latex{2x– y = 4} at its point that is common with the\latex{ x }-axis. How many solutions are there?

In an isosceles right-angled triangle the vertex at the right angle is the origin, and the equation of its inscribed circle is \latex{x(x-2)^{2}+(y-2)^{2}=4.} Calculate the coordinates of the other two vertices.

{{exercise_number}}. Give the equation of the straight line that touches the circle with the equation \latex{(x+2)^{2}+(y-5)^{2}=25,} and

is parallel with the straight line with the equation \latex{2x– 5y = –7;}

is perpendicular to the straight line with the equation \latex{2x– 5y = –7.}

How many solutions are there?

{{exercise_number}}. Calculate the coordinates of the intersection points of the circles with the equations \latex{(x+3)x^{2}+(y-4) ^{2}=16} and \latex{(x-4)^{2}+(y-2)^{2}=25. }

{{exercise_number}}. The equation of the circle \latex{k_{1}} is \latex{(x-3)^{2}+(y-6)^{2}=16,} the centre of the circle \latex{k_{2}} is \latex{K_{2}(-2;-3).} How long is the radius of \latex{k_{2}} if

the two circles have no points in common;

the two circles touch each other;

the two circles have two points in common?

{{exercise_number}}. Give the equation of the tangent line that can be drawn to the circle with the equation \latex{(x-3)^{2}+y^{2}=8} through the point \latex{ P }, if

\latex{P(-5;0);}

\latex{P(-2;4);}

\latex{P(8;1);}

\latex{P(2;-7);}

\latex{P(3;2).}

Mathematics 11.

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