Chapter 9: Coordinate Geometry of the Circle
9.1 Equation of a Circle: \((x - a)^2 + (y - b)^2 = r^2\)
Definition:
A circle is the set of all points \((x, y)\) that are at a fixed distance (the radius \(r\)) from a fixed point (the centre \((a, b)\)).
Standard form of the equation:
\[
(x - a)^2 + (y - b)^2 = r^2
\]
where:
- \((a, b)\) is the centre of the circle
- \(r\) is the radius (with \(r > 0\))
Proof: Derivation of the circle equation
Let \((x, y)\) be any point on the circle, and let \((a, b)\) be the centre.
By definition, the distance from \((x, y)\) to \((a, b)\) is \(r\):
\[
\sqrt{(x - a)^2 + (y - b)^2} = r
\]
Squaring both sides:
\[
(x - a)^2 + (y - b)^2 = r^2
\]
This is the Pythagorean theorem applied to the right-angled triangle formed by the horizontal distance \((x - a)\) and vertical distance \((y - b)\).
Special case:
If the centre is at the origin \((0, 0)\), the equation simplifies to:
\[
x^2 + y^2 = r^2
\]
General form (expanded):
\[
x^2 + y^2 + Dx + Ey + F = 0
\]
Where \(D = -2a\), \(E = -2b\), and \(F = a^2 + b^2 - r^2\).
This form is less useful for identifying centre and radius directly, but it can be converted back to standard form by completing the square.
Worked Example 9.1 (Writing the equation of a circle – non-calculator)
Write down the equation of a circle with:
- Centre \((3, -2)\) and radius \(5\)
- Centre \((-1, 4)\) and radius \(\sqrt{7}\)
Solution:
(a) \((x - 3)^2 + (y + 2)^2 = 25\)
(b) \((x + 1)^2 + (y - 4)^2 = 7\)
Worked Example 9.2 (Finding centre and radius from equation – non-calculator)
Find the centre and radius of the circle:
\[
(x + 2)^2 + (y - 5)^2 = 36
\]
Solution:
Compare with \((x - a)^2 + (y - b)^2 = r^2\):
\(x + 2 = x - (-2)\) so \(a = -2\)
\(y - 5 = y - 5\) so \(b = 5\)
\(r^2 = 36 \Rightarrow r = 6\)
Answer: Centre \((-2, 5)\), radius \(6\).
Tip: Remember that the numbers in the brackets have the opposite signs to the coordinates of the centre. For \((x + 2)\), the centre's x-coordinate is \(-2\).
Worked Example 9.3 (Centre at origin – non-calculator)
Write the equation of a circle with centre \((0, 0)\) and radius \(4\).
Solution:
\[
x^2 + y^2 = 16
\]
Worked Example 9.4 (Mauritian context – circular garden)
A circular garden in Port Louis has its centre at \((20, 15)\) on a map where units are in metres. The garden has a radius of 10 m. Find the equation of the circle.
Solution:
\[
(x - 20)^2 + (y - 15)^2 = 100
\]
9.2 Finding Centre and Radius from an Equation
When the equation is given in general form:
\[
x^2 + y^2 + Dx + Ey + F = 0
\]
We must complete the square to convert it to standard form.
Steps:
- Group \(x\) terms and \(y\) terms: \((x^2 + Dx) + (y^2 + Ey) + F = 0\)
- Complete the square for each:
\(x^2 + Dx = \left(x + \frac{D}{2}\right)^2 - \left(\frac{D}{2}\right)^2\)
\(y^2 + Ey = \left(y + \frac{E}{2}\right)^2 - \left(\frac{E}{2}\right)^2\)
- Rearrange to standard form and identify centre and radius.
Worked Example 9.5 (Completing the square – non-calculator)
Find the centre and radius of:
\[
x^2 + y^2 + 6x - 4y - 12 = 0
\]
Solution:
Group terms: \((x^2 + 6x) + (y^2 - 4y) = 12\)
Complete the square:
\[
x^2 + 6x = (x + 3)^2 - 9
\]
\[
y^2 - 4y = (y - 2)^2 - 4
\]
Substitute:
\[
(x + 3)^2 - 9 + (y - 2)^2 - 4 = 12
\]
\[
(x + 3)^2 + (y - 2)^2 = 12 + 9 + 4 = 25
\]
Answer: Centre \((-3, 2)\), radius \(5\).
Worked Example 9.6 (Mauritian context – circular reservoir)
A circular reservoir on a map has equation:
\[
x^2 + y^2 - 8x + 6y - 24 = 0
\]
Find its centre and radius.
Solution:
\((x^2 - 8x) + (y^2 + 6y) = 24\)
Complete the square:
\[
x^2 - 8x = (x - 4)^2 - 16
\]
\[
y^2 + 6y = (y + 3)^2 - 9
\]
Substitute:
\[
(x - 4)^2 - 16 + (y + 3)^2 - 9 = 24
\]
\[
(x - 4)^2 + (y + 3)^2 = 49
\]
Answer: Centre \((4, -3)\), radius \(7\).
Quick formulas from general form:
Centre: \(\left(-\frac{D}{2}, -\frac{E}{2}\right)\)
Radius: \(r = \sqrt{\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F}\)
9.3 Determining Whether a Point Lies Inside, On, or Outside a Circle
For a circle with centre \((a, b)\) and radius \(r\), a point \((x_1, y_1)\) lies:
- On the circle if: \((x_1 - a)^2 + (y_1 - b)^2 = r^2\)
- Inside the circle if: \((x_1 - a)^2 + (y_1 - b)^2 < r^2\)
- Outside the circle if: \((x_1 - a)^2 + (y_1 - b)^2 > r^2\)
Proof: Inside/outside condition
The expression \((x_1 - a)^2 + (y_1 - b)^2\) is the square of the distance from the point to the centre.
Comparing this to \(r^2\) determines whether the distance is less than, equal to, or greater than the radius.
Worked Example 9.7 (Point inside circle – non-calculator)
Does the point \((5, 2)\) lie inside, on, or outside the circle \((x - 3)^2 + (y + 1)^2 = 25\)?
Solution:
Centre: \((3, -1)\), radius: \(5\)
Distance squared from centre to point:
\[
(5 - 3)^2 + (2 - (-1))^2 = 2^2 + 3^2 = 4 + 9 = 13
\]
Since \(13 < 25\), the point lies inside the circle.
Worked Example 9.8 (Mauritian context – location check)
A circular park in Curepipe has centre \((10, 5)\) and radius \(8\) (units in km). A school is located at \((15, 9)\). Is the school inside the park?
Solution:
Distance squared from centre to school:
\[
(15 - 10)^2 + (9 - 5)^2 = 5^2 + 4^2 = 25 + 16 = 41
\]
Since \(41 < 64\), the school is inside the park.
9.4 Intersection of a Line and a Circle
To find the intersection of a line \(y = mx + c\) and a circle \((x - a)^2 + (y - b)^2 = r^2\):
- Substitute the line equation into the circle equation
- Expand and simplify to get a quadratic in \(x\) (or \(y\))
- Use the discriminant to determine the number of intersections:
| Discriminant |
Number of intersections |
| \(\Delta > 0\) |
Two points (line cuts the circle) |
| \(\Delta = 0\) |
One point (line is tangent to the circle) |
| \(\Delta < 0\) |
No intersection (line misses the circle) |
Proof: Discriminant determines intersection type
After substitution, the quadratic in \(x\) gives the x-coordinates of the intersection points. The number of real solutions is determined by the discriminant:
- \(\Delta > 0\): two distinct real x-values \(\rightarrow\) two points
- \(\Delta = 0\): one repeated x-value \(\rightarrow\) one point (tangent)
- \(\Delta < 0\): no real x-values \(\rightarrow\) no intersection
This follows directly from the quadratic formula.
Worked Example 9.9 (Line cuts circle – non-calculator)
Find the points of intersection of the line \(y = x + 1\) and the circle \((x - 2)^2 + (y - 3)^2 = 10\).
Solution:
Substitute \(y = x + 1\):
\[
(x - 2)^2 + (x + 1 - 3)^2 = 10
\]
\[
(x - 2)^2 + (x - 2)^2 = 10
\]
\[
2(x - 2)^2 = 10
\]
\[
(x - 2)^2 = 5
\]
\[
x - 2 = \pm \sqrt{5} \Rightarrow x = 2 \pm \sqrt{5}
\]
When \(x = 2 + \sqrt{5}\): \(y = 3 + \sqrt{5}\)
When \(x = 2 - \sqrt{5}\): \(y = 3 - \sqrt{5}\)
Answer: \((2 + \sqrt{5}, 3 + \sqrt{5})\) and \((2 - \sqrt{5}, 3 - \sqrt{5})\).
Worked Example 9.10 (Tangent to circle – non-calculator)
Find the value of \(k\) such that the line \(y = 2x + k\) is tangent to the circle \((x - 1)^2 + y^2 = 4\).
Solution:
Substitute \(y = 2x + k\):
\[
(x - 1)^2 + (2x + k)^2 = 4
\]
\[
x^2 - 2x + 1 + 4x^2 + 4kx + k^2 = 4
\]
\[
5x^2 + (4k - 2)x + (k^2 - 3) = 0
\]
For tangency, \(\Delta = 0\):
\[
(4k - 2)^2 - 4(5)(k^2 - 3) = 0
\]
\[
16k^2 - 16k + 4 - 20k^2 + 60 = 0
\]
\[
-4k^2 - 16k + 64 = 0
\]
\[
k^2 + 4k - 16 = 0
\]
\[
k = \frac{-4 \pm \sqrt{16 + 64}}{2} = \frac{-4 \pm \sqrt{80}}{2} = -2 \pm 2\sqrt{5}
\]
Answer: \(k = -2 + 2\sqrt{5}\) or \(k = -2 - 2\sqrt{5}\).
9.5 Tangents to a Circle
A tangent to a circle is a line that touches the circle at exactly one point. At the point of contact, the tangent is perpendicular to the radius.
Finding the equation of a tangent:
- Find the gradient of the radius from the centre to the point of contact
- The tangent gradient is the negative reciprocal
- Use point-gradient form to write the equation
Proof: Tangent is perpendicular to radius
The radius from the centre to the point of contact is the shortest distance from the centre to the line. For a line to be tangent, it must meet the circle at exactly one point, and the perpendicular from the centre to the line meets it at that point. Therefore, the radius is perpendicular to the tangent.
Worked Example 9.11 (Finding tangent – non-calculator)
Find the equation of the tangent to the circle \((x - 2)^2 + (y + 3)^2 = 25\) at the point \((6, 0)\).
Solution:
Centre: \((2, -3)\)
Gradient of radius:
\[
m_r = \frac{0 - (-3)}{6 - 2} = \frac{3}{4}
\]
Gradient of tangent (negative reciprocal):
\[
m_t = -\frac{4}{3}
\]
Equation through \((6, 0)\):
\[
y - 0 = -\frac{4}{3}(x - 6)
\]
\[
y = -\frac{4}{3}x + 8
\]
Answer: \(y = -\frac{4}{3}x + 8\).
Worked Example 9.12 (Mauritian context – circular path)
A circular path in a park has centre at \((4, 3)\) and radius \(5\). A runner is at point \((8, 6)\) on the path. Find the direction of the path (tangent) at this point.
Solution:
Gradient of radius from \((4, 3)\) to \((8, 6)\):
\[
m_r = \frac{6 - 3}{8 - 4} = \frac{3}{4}
\]
Gradient of tangent:
\[
m_t = -\frac{4}{3}
\]
Answer: The direction of the path is given by gradient \(-\frac{4}{3}\).
Chapter 9 Summary
| Concept |
Key points |
| Standard form |
\((x - a)^2 + (y - b)^2 = r^2\) |
| Centre |
\((a, b)\) (watch for opposite signs) |
| Radius |
\(r\) (positive square root) |
| General form |
\(x^2 + y^2 + Dx + Ey + F = 0\) |
| Completing the square |
Convert general form to standard form |
| Point location |
Compare distance squared to \(r^2\) |
| Line–circle intersection |
Substitute \(\rightarrow\) quadratic \(\rightarrow\) discriminant |
| Tangent |
Perpendicular to radius; \(m_1 \times m_2 = -1\) |
Exercises – Chapter 9
Easy (Drill – Non-Calculator)
- Write the equation of a circle with:
(a) Centre \((2, 3)\) and radius \(4\)
(b) Centre \((-1, 5)\) and radius \(6\)
(c) Centre \((0, 0)\) and radius \(\sqrt{10}\)
- Find the centre and radius of:
(a) \((x - 3)^2 + (y + 2)^2 = 16\)
(b) \((x + 4)^2 + (y - 1)^2 = 25\)
- Does the point \((4, 1)\) lie inside, on, or outside the circle \((x - 2)^2 + (y - 3)^2 = 9\)?
Medium (Examination Style)
- Find the centre and radius of: \(x^2 + y^2 - 6x + 4y - 12 = 0\)
- Mauritian context: A circular lake on a map has equation: \(x^2 + y^2 + 10x - 6y - 15 = 0\). Find its centre and radius.
- Find the points of intersection of the line \(y = 2x - 1\) and the circle \((x - 2)^2 + (y + 1)^2 = 20\).
- Find the value of \(k\) for which the line \(y = x + k\) is tangent to the circle \((x - 1)^2 + (y - 2)^2 = 8\).
Hard (Challenge for A*)
- Find the equation of the tangent to the circle \((x - 3)^2 + (y + 2)^2 = 25\) at the point \((6, 2)\).
- Mauritian context: A roundabout in Port Louis has centre at \((15, 10)\) and radius \(8\) (units in metres). A road follows the line \(y = \frac{3}{4}x + 4\).
(a) Does the road intersect the roundabout? Justify your answer.
(b) If it does, find the coordinates of the intersection points.
(c) If the road is to be tangent to the roundabout, find the new y-intercept of the road.
- The circle \(x^2 + y^2 = 25\) intersects the line \(y = mx + 5\) at two distinct points. Find the range of values of \(m\).