Equations of Circles: practice questions

6 questions · 12 marks · free, no account · Read the notes first

1. Question 1Multiple choice · 1 mark

   Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.

   A

   centre $(-3, 2)$, radius $5$

   B

   centre $(3, -2)$, radius $5$

   C

   centre $(3, -2)$, radius $12$

   D

   centre $(6, -4)$, radius $5$

2. Question 2Multiple choice · 1 mark

   Find the equation of the circle with centre $(1, -3)$ and radius 4.

   A

   $x^2 + y^2 + 2x - 6y - 6 = 0$

   B

   $x^2 + y^2 - 2x + 6y + 6 = 0$

   C

   $x^2 + y^2 - 2x + 6y - 6 = 0$

   D

   $x^2 + y^2 - 2x + 6y - 16 = 0$

3. Question 3Multiple choice · 1 mark

   Where is the point $(4, 1)$ relative to the circle $x^2 + y^2 - 4x + 2y - 4 = 0$?

   A

   Inside the circle

   B

   On the circle

   C

   Outside the circle

   D

   It cannot be determined

4. Question 4Multiple choice · 1 mark

   The straight line $y = x + k$ is a tangent to the circle $x^2 + y^2 = 8$. Find $k$.

   A

   $\pm 2\sqrt{2}$

   B

   $4$ only

   C

   $\pm 8$

   D

   $\pm 4$

5. Question 54 marks

   The circle $C$ is $x^2 + y^2 - 2x - 4y - 20 = 0$, and $P(4, 6)$ lies on $C$. Find the equation of the tangent to $C$ at $P$.

   Show worked solution

   The coordinates of the centre of $C$ are $(1, 2)$. 1A

   Slope of the radius through $P = \dfrac{6 - 2}{4 - 1} = \dfrac{4}{3}$ 1M

   So the slope of the tangent is $-\dfrac{3}{4}$ (tangent ⊥ radius): $y - 6 = -\dfrac{3}{4}(x - 4)$ 1M

   $3x + 4y - 36 = 0$ 1A

6. Question 64 marks

   The straight line $x - y + 1 = 0$ meets the circle $x^2 + y^2 = 13$ at $A$ and $B$. Find the coordinates of $A$ and $B$, and the length of $AB$.

   Show worked solution

   Substituting $y = x + 1$: $x^2 + (x + 1)^2 = 13$, i.e. $x^2 + x - 6 = 0$ 1M

   $x = 2$ or $x = -3$ 1A

   So $A(2, 3)$ and $B(-3, -2)$. 1A

   $AB = \sqrt{(2 + 3)^2 + (3 + 2)^2} = 5\sqrt{2}$ 1A