Equation of a circle: Practise Online

Either paper 2 marks

(a)
A circle has centre \(O\), the origin, and radius \(18\).

Write down the equation of the circle.

(b)
A different circle has equation \(x^2 + y^2 = 196\)

Write down the radius of this circle.
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(a) \(x^2 + y^2 = 324\)
(b) \(14\)
A circle centred on the origin with radius \(r\) has equation \(x^2 + y^2 = r^2\). The number on the right is the radius squared, not the radius.

\(18^2 = 324\), so the equation is \(x^2 + y^2 = 324\)
The number on the right is the radius squared, so the radius is its square root.

\(\sqrt{ 196 } = 14\)
Either paper 3 marks

The circle \(C\) has equation \(x^2 + y^2 = 9\)

(a)
Write down the coordinates of the two points where \(C\) crosses the \(y\)-axis.

(b)
State the coordinates of the point where \(C\) crosses the negative \(x\)-axis.
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(a) \((0, 3)\) and \((0, -3)\)
(b) \((-3, 0)\)
Every point on the \(y\)-axis has \(x = 0\), so put \(x = 0\) into the equation of the circle:

\(y^2 = 9\)

\(y = \sqrt{ 9 } = 3\) or \(y = -3\)

Both roots count — the circle crosses the \(y\)-axis on each side of the origin, at \((0, 3)\) and \((0, -3)\).
On the \(x\)-axis, \(y = 0\), so \(x^2 = 9\) and \(x = 3\) or \(x = -3\).

The negative \(x\)-axis is the half to the left of the origin, so the point is \((-3, 0)\).
- (a) M1puts \(x = 0\) to get \(y^2 = 9\)
- (a) A1\((0, 3)\) and \((0, -3)\)
Calculator 3 marks

A circle has centre the origin and equation \(x^2 + y^2 = 52\)

(a)
Calculate the radius of the circle.

Give your answer in the form \(a\sqrt{b}\), where \(a\) and \(b\) are integers and \(b\) is as small as possible.

(b)
Calculate the radius correct to 1 decimal place.
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(a) \(2\sqrt{ 13 }\)
(b) \(7.2\)
The number on the right is the radius squared, so the radius is \(\sqrt{ 52 }\), and \(52\) is not a square number.

Look for the largest square number that divides it:

\(52 = 4 \times 13\)

\(\sqrt{ 52 } = \sqrt{ 4 } \times \sqrt{ 13 } = 2\sqrt{ 13 }\)
\(2\sqrt{ 13 } \approx 7.2\)

Round from the exact value rather than from a value you have already rounded once.
- (a) M1\(52 = 4 \times 13\)
- (a) A1radius \(2\sqrt{ 13 }\)
Either paper 3 marks

The circle \(C\) has equation \(x^2 + y^2 = 400\)

Does the point \((-12, 17)\) lie inside \(C\), on \(C\), or outside \(C\)?

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Outside the circle
Work out \(x^2 + y^2\) for the point and compare it with \(400\). Less than \(400\) means inside, exactly \(400\) means on the circle, and more than \(400\) means outside.

\((-12)^2 + (17)^2 = 144 + 289 = 433\)

\(433\) is greater than \(400\), so the point lies outside the circle.
- M2work out \(x^2 + y^2\): \((-12)^2 + (17)^2 = 144 + 289 = 433\)
- A1outside the circle, comparing \(433\) with \(400\)
Either paper 3 marks

A circle has centre \(O\), the origin, and passes through the point \((-12, 5)\).

(a)
Find the equation of the circle.

(b)
Write down the radius of the circle.
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(a) \(x^2 + y^2 = 169\)
(b) \(13\)
The equation is \(x^2 + y^2 = r^2\) for some \(r\), and the point on the circle has to fit it. Put its coordinates in:

\((-12)^2 + (5)^2 = 144 + 25 = 169\)

So \(r^2 = 169\) and the equation is \(x^2 + y^2 = 169\)
\(r^2 = 169\), so \(r = \sqrt{ 169 } = 13\)
- (a) M1\((-12)^2 + (5)^2 = 144 + 25\)
- (a) A1\(x^2 + y^2 = 169\)
Either paper 7 marks

\(A\), \(B\) and \(C\) are points on the circle \(x^2 + y^2 = 64\). The circle is not drawn; the points are shown on the axes.

\(A\) is on the \(y\)-axis.

\(B\) is on the \(x\)-axis.

\(M\) is the midpoint of \(AB\).

\(COM\) is a straight line.

(a)
Show that the coordinates of \(A\) are \((0, 8)\).

(b)
Work out the coordinates of \(B\).

(c)
Show that the equation of the straight line passing through \(C\), \(O\) and \(M\) is \(y = -x\).

(d)
Work out the coordinates of \(C\).

Give your answer in surd form.
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(a) \(y^2 = 64\), so \(y = 8\) (above \(O\)): \(A = (0, 8)\)
(b) \((-8, 0)\)
(c) \(M = (-4, 4)\), gradient \(-1\), through \(O\): \(y = -x\)
(d) \((4\sqrt{2}, -4\sqrt{2})\)
\(A\) is on the \(y\)-axis, so \(x = 0\).

Substitute into the circle: \(0^2 + y^2 = 64\), so \(y^2 = 64\).

\(y = \sqrt{ 64 } = 8\), taking the positive value because \(A\) is above \(O\).

\(A = (0, 8)\)
\(B\) is on the \(x\)-axis, so \(y = 0\), and \(x^2 + 0^2 = 64\) gives \(x^2 = 64\), \(x = \pm 8\).

In the diagram \(B\) is on the side of \(O\) with \(x\) negative, so \(B = (-8, 0)\).
\(A = (0, 8)\) and \(B = (-8, 0)\), so

\(M = \left( \frac{ 0 - 8 }{2}, \frac{ 8 + 0 }{2} \right) = (-4, 4)\)

The line through \(O\) and \(M\) has gradient \(\frac{ 4 - 0 }{ -4 - 0 } = -1\).

It passes through the origin, so \(c = 0\) and the equation is \(y = -x\).
\(C\) is on the line \(y = -x\), so \(y^2 = x^2\). Substitute into \(x^2 + y^2 = 64\):

\(x^2 + x^2 = 64\), so \(2x^2 = 64\) and \(x^2 = 32\).

\(x = \pm\sqrt{ 32 } = \pm\sqrt{ 2 \times 16 } = \pm 4\sqrt{2}\)

\(M\) is at \(x = -4\), and \(C\) is on the other side of \(O\) from \(M\) (\(COM\) is a straight line), so \(C\) has \(y = -4\sqrt{2}\) and

\(C = (4\sqrt{2}, -4\sqrt{2})\)
- (c) M1midpoint \(M = (-4, 4)\)
- (c) A1gradient of \(OM\) is \(-1\) and the line passes through \(O\), so \(y = -x\)
- (d) M1substitute \(y = -x\) into the circle: \(2x^2 = 64\)
- (d) M1\(x^2 = 32\), so \(x = \pm 4\sqrt{2}\)
- (d) A1C is on the opposite side of O from M: \((4\sqrt{2}, -4\sqrt{2})\)
Either paper 5 marks

(a)
On the grid, draw the graph of \(x^2 + y^2 = 64\)

On the same grid, draw the graph of \(6y = -x\)

(b)
Use your graphs to find estimates for the solutions of the simultaneous equations

\(x^2 + y^2 = 64\)

\(6y = -x\)
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You cannot draw on this page. Draw (a) on paper, then show the answer and check your drawing against the model.
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(a) A circle with centre \(O\) (the origin) and radius \(8\), and the straight line through \(O\) and \((6, -1)\)
(b) About \((7.9, -1.3)\) and \((-7.9, 1.3)\) (each coordinate within 0.3)
(a)

The equation \(x^2 + y^2 = 64\) is a circle with centre \(O\), the origin.

Its radius is \(\sqrt{ 64 } = 8\), so the circle passes through \((8, 0)\), \((0, 8)\), \((-8, 0)\) and \((0, -8)\).

Rearrange \(6y = -x\) to \(y = -\frac{ 1 }{ 6 }x\).

This is a straight line through \(O\). For every \(6\) squares across it goes down \(1\), so it also passes through \((6, -1)\).
The solutions are the two points where the line crosses the circle.

Reading them off the graph gives about \((7.9, -1.3)\) and \((-7.9, 1.3)\).

The line passes through \(O\), the centre, so the two crossings are opposite ends of a diameter. The second is the first with both signs reversed.
- (a) B2the circle with centre \(O\) and radius \(\sqrt{ 64 } = 8\) drawn
- (a) B1the circle, and the straight line through \(O\) with gradient \(-\frac{ 1 }{ 6 }\), and nothing else
- (b) A1one crossing read as about \((7.9, -1.3)\)
- (b) A1other crossing read as about \((-7.9, 1.3)\)
Either paper 3 marks

The point \((15, k)\) lies on the circle \(x^2 + y^2 = 625\)

Given that \(k\) is positive, work out the value of \(k\).
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\(20\)
A point on the circle fits its equation, so put \(x = 15\) and \(y = k\) into it.

\((15)^2 + k^2 = 625\)

\(225 + k^2 = 625\)

\(k^2 = 400\)

A square root has two values: \(k = 20\) or \(k = -20\). The question says \(k\) is positive, so \(k = 20\).
- M1substitutes \(x = 15\) into the circle equation: \((15)^2 + k^2 = 625\)
- M1rearranges to \(k^2 = 400\)
- A1chooses the positive root, \(k = 20\)
Calculator 4 marks

The diagram shows a circle, centre \(O\), drawn on coordinate axes with \(O\) at the origin.

\(BA\) is the tangent to the circle at the point \(A\).

Angle \(OBA = 52^\circ\)

Point \(B\) has coordinates \((27, 0)\).

Point \(P\) has coordinates \((2p, p)\) and lies on the circle, where \(p > 0\).

Work out the value of \(p\).

Give your answer correct to 1 decimal place.

You must show all your working.
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\(9.5\)
The radius \(OA\) meets the tangent \(BA\) at a right angle, so triangle \(OAB\) is right-angled at \(A\), with hypotenuse \(OB = 27\).

\(r = OA = 27 \sin 52^\circ \approx 21.28\)

The circle is centred on the origin, so its equation is

\(x^2 + y^2 = r^2 \approx 452.68\)

Point \(P\) lies on the circle, so put \(x = 2p\) and \(y = p\) into the equation:

\((2p)^2 + p^2 \approx 452.68\)

\(4p^2 + p^2 \approx 452.68\)

\(5p^2 \approx 452.68\)

\(p^2 \approx \frac{ 452.68 }{ 5 }\)

Since \(p > 0\), \(p \approx 9.52\), so \(p \approx 9.5\) to 1 decimal place.
- M1radius OA is perpendicular to the tangent, so in triangle OAB r = 27 sin 52° ≈ 21.28
- M1circle centred on the origin: x^2 + y^2 = r^2 ≈ 452.68
- M1substitute P: 4p^2 + p^2 ≈ 452.68, so 5p^2 ≈ 452.68
- A1p ≈ 9.52, so p = 9.5 to 1 d.p. (taking the positive root)
Either paper 4 marks

A circle has equation \(x^2 + y^2 = 100\)

The point \(P(-6, 8)\) lies on the circle.

Nadia says that the tangent to the circle at \(P\) crosses the \(y\)-axis at the point \((0, 16)\).

Is Nadia correct?

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No: the tangent crosses the \(y\)-axis at \((0, 12.5)\)
Build the tangent, find where it meets the axis, then compare.

The gradient of the radius \(OP\) is \(\frac{ 8 - 0 }{ -6 - 0 } = -\frac{4}{3}\)

The tangent meets the radius at right angles, so its gradient is

\(-1 \div (-\frac{4}{3}) = \frac{3}{4}\)

The tangent passes through \(P(-6, 8)\), so

\(y - 8 = \frac{3}{4}(x + 6)\)

The \(y\)-axis is where \(x = 0\):

\(y - 8 = \frac{3}{4} \times (6) = \frac{9}{2}\)

\(y = 8 + \frac{9}{2} = \frac{25}{2}\)

The \(x\)-axis is where \(y = 0\):

\(-8 = \frac{3}{4}(x + 6)\)

\(x + 6 = -8 \div \frac{3}{4} = -\frac{32}{3}\)

\(x = -6 - \frac{32}{3} = -\frac{50}{3}\)

Nadia is asking about the \(y\)-axis, where the tangent crosses at \((0, 12.5)\).

Nadia gave \((0, 16)\), which is not that point.

Answer: No
- M1gradient of \(OP\), \(\frac{ 8 - 0 }{ -6 - 0 } = -\frac{4}{3}\)
- M1gradient of the tangent, \(\frac{3}{4}\)
- M1tangent through \(P\), \(y - 8 = \frac{3}{4}(x + 6)\)
- A1No: crossing \((0, 12.5)\) against the claimed \((0, 16)\)