3.3 The equation of a circle and its properties — Practise with help

Work each one first, closed book. Decide at the start whether the question has given you the centre and the radius, or an expanded equation that hides them.

Warm-up: remember

Write $x^2 - 10x + 7$ in completed-square form.
Work out the distance between $(1, 2)$ and $(4, 6)$ .
Work out the gradient of the line $3x + 4y = 12$ .

Answers

Half of $-10$ is $-5$ : $(x - 5)^2 - 25 + 7 = (x - 5)^2 - 18$ .
$\sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .
$4y = 12 - 3x$ , so $y = 3 - \dfrac{3}{4}x$ . The gradient is $-\dfrac{3}{4}$ .

If question 1 was hard, revisit completing the square in Section 2.3 before you continue.

Practise with help

1. Write the equation of the circle with centre $(2, -5)$ and radius $6$ .

Hint

Substitute the centre and the radius into $(x - a)^2 + (y - b)^2 = r^2$ .

Next step

The centre gives $a = 2$ and $b = -5$ . Subtracting a negative gives a plus on the left: $(y - (-5))^2 = (y + 5)^2$ .

Full working

(x - 2)^2 + (y - (-5))^2 = 6^2,

so the equation is

(x - 2)^2 + (y + 5)^2 = 36.

The radius is $6$ , and $6^2 = 36$ is the number on the right.

2. Find the centre and the radius of $x^2 + y^2 - 8x + 2y - 8 = 0$ .

Hint

Complete the square twice, once in $x$ and once in $y$ . Remember to subtract the number you add.

Next step

$x^2 - 8x = (x - 4)^2 - 16$ and $y^2 + 2y = (y + 1)^2 - 1$ . Put those into the equation and collect the constants.

Full working

(x - 4)^2 - 16 + (y + 1)^2 - 1 - 8 = 0.

Add the constants:

(x - 4)^2 + (y + 1)^2 - 25 = 0, \qquad (x - 4)^2 + (y + 1)^2 = 25.

The centre is $(4, -1)$ and the radius is $5$ .

3. The circle is $(x - 2)^2 + (y - 1)^2 = 25$ . Decide whether the point $(7, 3)$ is inside, on, or outside the circle.

Hint

Substitute the point into the left side and compare the answer with $25$ .

Next step

Work out $(7 - 2)^2 + (3 - 1)^2$ , then compare with $r^2 = 25$ .

Full working

(7 - 2)^2 + (3 - 1)^2 = 25 + 4 = 29.

Since $29 > 25$ , the point is outside the circle. The distance from the centre is $\sqrt{29} \approx 5.39$ , which is greater than the radius $5$ .

4. Find the tangent to the circle $x^2 + y^2 = 169$ at the point $(5, 12)$ .

Hint

Find the gradient of the radius from the centre to the point. The tangent has the negative reciprocal gradient.

Next step

The radius has gradient $\frac{12}{5}$ , so the tangent has gradient $-\frac{5}{12}$ . Use the point $(5, 12)$ .

Full working

y - 12 = -\frac{5}{12}(x - 5).

Multiply by $12$ :

12y - 144 = -5x + 25, \qquad 5x + 12y = 169.

Check the point: $5 \times 5 + 12 \times 12 = 25 + 144 = 169$ .

5. Find where the line $y = 3$ meets the circle $x^2 + y^2 = 25$ .

Hint

Substitute $y = 3$ into the circle equation. The result is a quadratic in $x$ .

Next step

$x^2 + 9 = 25$ , so $x^2 = 16$ . Take both square roots.

Full working

x^2 + 3^2 = 25, \qquad x^2 + 9 = 25, \qquad x^2 = 16, \qquad x = 4 \text{ or } x = -4.

The line meets the circle at $(4, 3)$ and $(-4, 3)$ . The chord is horizontal and $8$ units long.

6. The line $x + y = 19$ cuts the circle $(x - 6)^2 + (y - 8)^2 = 25$ at $(11, 8)$ and $(6, 13)$ . Work out the length of the chord.

Hint

The chord length is the distance between the two points. Use the distance formula.

Next step

Work out $\sqrt{(11 - 6)^2 + (8 - 13)^2}$ .

Full working

\sqrt{(11 - 6)^2 + (8 - 13)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}.

The chord is $5\sqrt{2} \approx 7.07$ units long. The midpoint is $\left(\frac{17}{2}, \frac{21}{2}\right)$ , and the perpendicular from the centre $(6, 8)$ passes through it.

7. 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.

Hint

$AB$ is a diameter. For a right angle, the two gradients $CA$ and $CB$ must multiply to $-1$ .

Next step

The gradient of $CA$ is $\dfrac{4 - 0}{-3 - (-5)} = \dfrac{4}{2} = 2$ . Work out the gradient of $CB$ the same way.

Full working

m_{CA} = \frac{4 - 0}{-3 - (-5)} = \frac{4}{2} = 2,

m_{CB} = \frac{4 - 0}{-3 - 5} = \frac{4}{-8} = -\frac{1}{2}.

The product is $2 \times \left(-\frac{1}{2}\right) = -1$ , so the two lines meet at a right angle. This is the angle in a semicircle: $AB$ is a diameter, and any point $C$ on the circle gives a right angle at $C$ .