Circles
Coordinate geometry
- AS
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Write down the equation of the circle with centre $(2, -3)$ and radius $5$.
3 marks
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Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 3 = 0$.
3 marks
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Find the points where the line $y = x + 1$ meets the circle $x^2 + y^2 = 25$.
4 marks
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The circle $x^2 + y^2 - 8x - 2y + 7 = 0$ has centre $C$ and radius $r$. Find $C$ and $r$, and decide whether the line $x + y = 7$ is a tangent to the circle.
4 marks
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Find the equation of the tangent to the circle $x^2 + y^2 = 25$ at the point $(3, 4)$.
5 marks
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The circle $(x - 1)^2 + (y - 2)^2 = 25$ meets the line $y = x + 1$ at two points. Find those points and the length of the chord joining them.
5 marks
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A circle has centre $(4, 1)$ and passes through $A(1, 5)$. Find the equation of the circle and the equation of the tangent at $A$.
6 marks
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The circle $(x - 4)^2 + (y - 1)^2 = 25$ has a tangent at $A(1, 5)$, found in the previous style of question, but this time the point of contact is $B(8, 4)$. Find the equation of the tangent at $B$, and show that this tangent is perpendicular to the tangent at $A(1, 5)$.
7 marks
Mark scheme
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Answer. $(x - 2)^2 + (y + 3)^2 = 25$
- Use $(x - a)^2 + (y - b)^2 = r^2$ with centre $(2, -3)$ and $r = 5$.
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Answer. The centre is $(3, -2)$ and the radius is $4$.
- Complete the square: $(x - 3)^2 - 9 + (y + 2)^2 - 4 - 3 = 0$.
- $(x - 3)^2 + (y + 2)^2 = 16$.
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Answer. The points are $(3, 4)$ and $(-4, -3)$.
- Substitute: $x^2 + (x + 1)^2 = 25$.
- $x^2 + x - 12 = 0$, so $(x + 4)(x - 3) = 0$.
- The points are $(3, 4)$ and $(-4, -3)$.
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Answer. The centre is $(4, 1)$ and the radius is $\sqrt{10}$. The line is not a tangent.
- $(x - 4)^2 + (y - 1)^2 = 10$.
- The distance from the centre to the line is $\dfrac{|4 + 1 - 7|}{\sqrt{2}} = \sqrt{2}$.
- $\sqrt{2} < \sqrt{10}$, so the line is a chord, not a tangent.
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Answer. $3x + 4y = 25$
- The radius from the origin to $(3, 4)$ has gradient $\dfrac{4}{3}$.
- The tangent is perpendicular to the radius, so its gradient is $-\dfrac{3}{4}$.
- The tangent at $(x_1, y_1)$ on $x^2 + y^2 = r^2$ is $xx_1 + yy_1 = r^2$, which gives $3x + 4y = 25$.
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Answer. The points are $\left(1 \pm \dfrac{5\sqrt{2}}{2},\, 2 \pm \dfrac{5\sqrt{2}}{2}\right)$, and the chord has length $10$.
- $(x - 1)^2 + (x - 1)^2 = 25$, so $(x - 1)^2 = \dfrac{25}{2}$.
- $x = 1 \pm \dfrac{5\sqrt{2}}{2}$ and $y = x + 1$.
- The differences in $x$ and in $y$ are both $5\sqrt{2}$.
- Length $= \sqrt{50 + 50} = 10$.
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Answer. $(x - 4)^2 + (y - 1)^2 = 25$, and the tangent is $y - 5 = \dfrac{3}{4}(x - 1)$.
- Radius squared $= 3^2 + 4^2 = 25$.
- The radius $CA$ has gradient $-\dfrac{4}{3}$.
- The tangent gradient is $\dfrac{3}{4}$.
- $y - 5 = \dfrac{3}{4}(x - 1)$.
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Answer. The tangent at $B$ is $y - 4 = -\dfrac{4}{3}(x - 8)$. It is perpendicular to the tangent at $A$.
- Check $B$ lies on the circle: $4^2 + 3^2 = 25$.
- Gradient of the radius to $B$ is $\dfrac{3}{4}$, so the tangent gradient is $-\dfrac{4}{3}$.
- The tangent at $A$ has gradient $\dfrac{3}{4}$, from the radius gradient $-\dfrac{4}{3}$.
- The product of the tangent gradients is $-1$, so the tangents are perpendicular.