3.3 The equation of a circle and its properties — Practise

Work through these on your own. For an expanded equation, complete the square before you do anything else: the centre and the radius decide every later step. For a tangent, remember that the radius meets it at a right angle. Later items reach back to gradients and simultaneous equations from Chapter 2, so decide which tool each one needs before you start.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Write the equation of the circle with centre $(3, -2)$ and radius $4$ , in expanded form.
Find the centre and the radius of the circle $x^2 + y^2 - 12x - 16y + 75 = 0$ .
Find the centre and the radius of the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ .
Decide whether the point $(5, 4)$ is inside, on, or outside the circle $(x - 1)^2 + (y - 2)^2 = 25$ .
A circle has centre $(2, 3)$ and passes through $(6, 6)$ . Find its equation.
Find the tangent to the circle $x^2 + y^2 = 169$ at the point $(5, 12)$ . Give the equation and show that the distance from the centre to your line equals the radius.
Find the two points where the line $y = 3$ meets the circle $x^2 + y^2 = 25$ .
Find the two points where the line $x + y = 19$ meets the circle $(x - 6)^2 + (y - 8)^2 = 25$ .
The line $x + y = 19$ cuts the circle $(x - 6)^2 + (y - 8)^2 = 25$ at $(6, 13)$ and $(11, 8)$ . Work out the length of the chord.
For the chord of $(x - 6)^2 + (y - 8)^2 = 25$ on the line $x + y = 19$ , with endpoints $(6, 13)$ and $(11, 8)$ , show that the perpendicular from the centre passes through the midpoint of the chord.
The circle $x^2 + y^2 = 25$ has $A(-5, 0)$ , $B(5, 0)$ and $C(-3, 4)$ on it. Show that angle $ACB$ is a right angle.
Show that the line $y = 2x + 5$ is a tangent to the circle $(x - 1)^2 + (y - 2)^2 = 5$ by comparing the distance from the centre to the line with the radius.
Decide whether the line $y = 10$ meets the circle $x^2 + y^2 = 25$ . Use the discriminant of the quadratic you get.

Answers and concise solutions

1. The equation

Answer: Substitute the centre and the radius into $(x - a)^2 + (y - b)^2 = r^2$ : $(x - 3)^2 + (y + 2)^2 = 16$ . Expanding gives $x^2 - 6x + 9 + y^2 + 4y + 4 = 16$ , so $x^2 + y^2 - 6x + 4y - 3 = 0$ . The expanded form is the same circle, written differently.

2. The centre and the radius

Answer: Complete the square twice: $(x - 6)^2 - 36 + (y - 8)^2 - 64 + 75 = 0$ , so $(x - 6)^2 + (y - 8)^2 = 25$ . The centre is $(6, 8)$ and the radius is $5$ . The centre is $\left(-\frac{D}{2}, -\frac{E}{2}\right)$ and the radius satisfies $r^2 = a^2 + b^2 - F$ .

3. The centre and the radius

Answer: Complete the square: $(x + 2)^2 - 4 + (y - 3)^2 - 9 - 12 = 0$ , so $(x + 2)^2 + (y - 3)^2 = 25$ . The centre is $(-2, 3)$ and the radius is $5$ . The signs matter: $(x + 2)^2$ means the centre coordinate is $-2$ , not $2$ .

4. The decision

Answer: Substitute the point: $(5 - 1)^2 + (4 - 2)^2 = 16 + 4 = 20$ . Since $20 < 25$ , the point is inside the circle. The distance from the centre is $\sqrt{20} \approx 4.47$ , which is less than the radius $5$ .

5. The equation

Answer: The radius is the distance from the centre to the point: $r^2 = (6 - 2)^2 + (6 - 3)^2 = 16 + 9 = 25$ , so $r = 5$ . The equation is $(x - 2)^2 + (y - 3)^2 = 25$ . Working with $r^2$ first avoids a square root, and the square root is only needed if the question asks for the radius itself.

6. The tangent and the distance check

Answer: The radius from $(0, 0)$ to $(5, 12)$ has gradient $\frac{12}{5}$ , so the tangent has gradient $-\frac{5}{12}$ : $y - 12 = -\frac{5}{12}(x - 5)$ , which is $5x + 12y = 169$ . The point satisfies the line, since $25 + 144 = 169$ . The distance from the centre $(0, 0)$ to the line is $\frac{|-169|}{\sqrt{5^2 + 12^2}} = \frac{169}{13} = 13$ , which equals the radius. A line at the radius from the centre touches the circle once, so it is a tangent.

7. The two points

Answer: Substitute $y = 3$ : $x^2 + 9 = 25$ , so $x^2 = 16$ and $x = 4$ or $x = -4$ . The two points are $(4, 3)$ and $(-4, 3)$ . The line meets the circle twice, so it is a secant, and the chord is $8$ units long.

8. The two points

Answer: Substitute $y = 19 - x$ : $(x - 6)^2 + (11 - x)^2 = 25$ , so $x^2 - 17x + 66 = 0$ and $(x - 6)(x - 11) = 0$ . Then $x = 6$ or $x = 11$ , giving $(6, 13)$ and $(11, 8)$ . Two distinct real roots mean the line cuts the circle twice.

9. The chord length

Answer: The chord length is the distance between the endpoints: $\sqrt{(11 - 6)^2 + (8 - 13)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$ , about $7.07$ . The exact surd is the answer; a rounded decimal is not.

10. The midpoint and the perpendicular

Answer: The midpoint is $\left(\frac{6 + 11}{2}, \frac{13 + 8}{2}\right) = \left(\frac{17}{2}, \frac{21}{2}\right)$ . The chord has gradient $-1$ , so the perpendicular from the centre $(6, 8)$ has gradient $1$ : $y = x + 2$ . Substituting the midpoint gives $\frac{21}{2} = \frac{17}{2} + 2$ , which is true. So the perpendicular from the centre meets the chord at its midpoint and bisects it.

11. The gradient product

Answer: The gradient of $CA$ is $\frac{4 - 0}{-3 - (-5)} = 2$ and the gradient of $CB$ is $\frac{4 - 0}{-3 - 5} = -\frac{1}{2}$ . Their product is $-1$ , so the two lines meet at a right angle. $AB$ is a diameter because it joins $(-5, 0)$ and $(5, 0)$ through the centre, so this is the angle in a semicircle.

12. The distance and the conclusion

Answer: Write the line as $2x - y + 5 = 0$ . The centre is $(1, 2)$ , so the distance is $\frac{|2 - 2 + 5|}{\sqrt{2^2 + 1^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$ . The radius is also $\sqrt{5}$ , since $r^2 = 5$ . The distance equals the radius, so the line touches the circle at exactly one point and is a tangent.

13. The decision and the discriminant

Answer: Substitute $y = 10$ : $x^2 + 100 = 25$ , so $x^2 + 75 = 0$ . The discriminant is $0^2 - 4 \times 1 \times 75 = -300$ , which is negative. A negative discriminant means no real roots, so the line misses the circle. This matches the distance check: the distance from $(0, 0)$ to $y = 10$ is $10$ , which is greater than the radius $5$ .