Families of circles and locus problems

What you will be able to do

• write the family of circles through the meeting points of a circle and a line, or of two circles;
• find the member of a family that satisfies one more condition;
• find a circle that touches a line at a given point;
• solve the standard locus problems about circles.

Pegs on a board

Push two pegs into a board. Many circles can be drawn through both pegs. Their centres all lie on the perpendicular bisector of the line between the pegs. If you also fix a third point, only one circle is left. A family of circles is the set of all such circles, with a number $\lambda$ that picks one member.

A circle and a line

Let $S = x^2 + y^2 + 2gx + 2fy + c = 0$ be a circle and $L = ax + by + d = 0$ a line. Every circle through the points where they meet has the equation

Why does this work? Any point on both $S = 0$ and $L = 0$ gives $0 + \lambda \times 0 = 0$, so it satisfies the equation. The equation is $x^2 + y^2 + (\text{terms of degree 1}) = 0$, so it is a circle. One value of $\lambda$ is fixed by one more condition. That condition may be a point on the circle, a line that the circle touches, or a line that holds the centre.

Example. Find the circle through the points where $x^2 + y^2 = 25$ meets $x + y = 1$ and through the origin. Put $(0, 0)$ into $x^2 + y^2 - 25 + \lambda(x + y - 1) = 0$: $-25 - \lambda = 0$, so $\lambda = -25$. The circle is $x^2 + y^2 - 25x - 25y = 0$. Its centre is $(12.5, 12.5)$ and its radius is $\sqrt{2 \times 156.25} \approx 17.68$. The two fixed points are $(4, -3)$ and $(-3, 4)$. Test one: $16 + 9 - 100 + 75 = 0$.

Two circles

For two circles $S_1 = 0$ and $S_2 = 0$ the family through their meeting points is

For $\lambda = -1$ the $x^2 + y^2$ terms cancel and you get the radical axis, which is the line through the meeting points. So the family contains one line and many circles.

A circle that touches a line at a point

A circle that touches the line $L = 0$ at the point $A(x_1, y_1)$ on it is a member of the family

The first part is a "circle of radius zero" at $A$, and $L = 0$ is the tangent at $A$. The two meet at $A$ with a double point, and that is what touching means.

Example. Find the circle that touches $3x + 4y - 10 = 0$ at $(2, 1)$ and passes through $(0, 3)$. First check that $(2, 1)$ is on the line: $6 + 4 - 10 = 0$. Put $(0, 3)$ into $(x - 2)^2 + (y - 1)^2 + \lambda(3x + 4y - 10) = 0$: $4 + 4 + \lambda(2) = 0$, so $\lambda = -4$. The circle is $x^2 + y^2 - 16x - 18y + 45 = 0$, with centre $(8, 9)$ and radius $10$. As a check, the distance of $(8, 9)$ from the line is $\dfrac{|24 + 36 - 10|}{5} = 10$.

Standard locus problems

A locus is the path of a point that obeys a rule. The method is always the same.

1. Call the moving point $(h, k)$.
2. Write the rule as an equation in $h$ and $k$.
3. Remove any extra letters. This is the part that needs thought.
4. Replace $(h, k)$ by $(x, y)$ and name the curve.

You met locus questions in the chapter on straight lines. Here are the standard ones with circles.

The Apollonius circle. Let $A$ and $B$ be fixed points. The locus of $P$ with $PA : PB = m : n$ is a circle when $m \ne n$ (and the perpendicular bisector when $m = n$). With $A(0, 0)$, $B(6, 0)$ and $PA = 2PB$, we get $x^2 + y^2 = 4\left[(x - 6)^2 + y^2\right]$, so $x^2 + y^2 - 16x + 48 = 0$. This is the circle with centre $(8, 0)$ and radius $4$. It meets the $x$-axis at $(4, 0)$, which divides $AB$ internally in $2 : 1$, and at $(12, 0)$, which divides it externally in $2 : 1$.

Midpoints of chords of fixed length. A chord of length $2l$ in $x^2 + y^2 = a^2$ has its midpoint at distance $\sqrt{a^2 - l^2}$ from the centre. So the midpoints lie on $x^2 + y^2 = a^2 - l^2$. For chords of length $6$ in $x^2 + y^2 = 25$ the locus is $x^2 + y^2 = 16$.

Equal tangent lengths. The point whose tangent lengths to two circles are equal moves on their radical axis. If the lengths are in a fixed ratio other than $1 : 1$, the locus is a circle.

Circles that touch two circles. Let a circle of radius $\rho$ and centre $P$ touch the circle $(C_1, r_1)$ and the circle $(C_2, r_2)$, both from outside. Then $PC_1 = \rho + r_1$ and $PC_2 = \rho + r_2$. The difference $PC_2 - PC_1 = r_2 - r_1$ does not depend on $\rho$. So $P$ moves on a branch of a hyperbola with foci $C_1$ and $C_2$. If the moving circle touches one of them from outside and the other from inside, the sum of the two distances is constant, and $P$ moves on an ellipse. The same idea holds for a circle through a fixed point that touches a fixed line: the centre is as far from the point as from the line, so it moves on a parabola. All three curves come back in the last courses of this chapter.

For example, let a circle touch $x^2 + y^2 = 1$ and $(x - 6)^2 + y^2 = 9$ from outside. Then $PC_2 - PC_1 = 3 - 1 = 2$. When the moving circle has radius $2$, we get $PC_1 = 3$ and $PC_2 = 5$, so $x^2 + y^2 = 9$ and $(x - 6)^2 + y^2 = 25$. Subtracting gives $-12x + 36 = 16$, so $x = \dfrac{5}{3}$ and $y = \pm\dfrac{2\sqrt{14}}{3} \approx \pm 2.49$.

Practise with new numbers

Try it in Python

The program steps through the family $S + \lambda L = 0$ for the circle and the line of the example. For each value of $\lambda$ it prints the centre and the radius and checks that the two fixed points are on the circle.

Common mistakes

• Using $S + \lambda L$ when $L$ is not the line through the meeting points. It must be the line that you intend to share.
• Putting $\lambda = -1$ for two circles and still calling it a circle. The result is a line.
• Forgetting to check that a point is on the tangent line. The family for a touching circle needs the point on the line.
• Losing a branch. A locus such as $PA = 2PB$ is the whole circle. Do not keep only the upper half.

Remember these

• $S + \lambda L = 0$ is the family of circles through the meeting points of the circle $S = 0$ and the line $L = 0$.
• $S_1 + \lambda S_2 = 0$ ($\lambda \ne -1$) is the family through the meeting points of two circles.
• A circle touching $L = 0$ at $(x_1, y_1)$ is $(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$.
• To find a locus: name $(h, k)$, write the rule, remove the extra letters, replace by $(x, y)$.
• A constant difference of distances gives a hyperbola, a constant sum an ellipse, and a point equal in distance to a point and a line a parabola.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the circle through the points where $x^2 + y^2 = 4$ meets $x = 1$ and through the origin.

2. In the family $S_1 + \lambda S_2 = 0$, what does $\lambda = -1$ give?

3. Find the locus of the midpoint of a chord of length $6$ in the circle $x^2 + y^2 = 25$.

4. Find the locus of $P$ such that $PA = 2PB$ for $A(0, 0)$ and $B(6, 0)$.

5. A circle touches $x^2 + y^2 = 1$ and $(x - 6)^2 + y^2 = 9$ from outside. What is the difference of the distances of its centre from the two centres?

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections). The families of circles and the locus problems are in the JEE Main syllabus, not in the NCERT text.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry).