Every point on a circle is the same distance from its centre. Write that distance with Pythagoras and you get the equation \((x - a)^{2} + (y - b)^{2} = r^{2}\). This lesson builds the equation from a centre and radius, a point or a diameter, reads the centre and radius back from any form, and finds the circle through three points.
The circle with centre \((a, b)\) and radius \(r\) has equation \((x - a)^{2} + (y - b)^{2} = r^{2}\). Expanded, a circle is \(x^{2} + y^{2} + 2gx + 2fy + c = 0\), with centre \((-g, -f)\) and radius \(\sqrt{g^{2} + f^{2} - c}\) (in the List of formulas); it is a real circle only when \(g^{2} + f^{2} - c > 0\). A point is inside, on or outside the circle as its distance squared from the centre is less than, equal to or greater than \(r^{2}\).
Fig. 1: a circle with centre C(a, b) and radius r. For any point P(x, y) on the circle, the horizontal side of the right-angled triangle is x - a, the vertical side is y - b and the hypotenuse is the radius r, so Pythagoras gives (x - a)² + (y - b)² = r²
Section 01
The equation is Pythagoras.
Take a circle with centre \(C(a, b)\) and radius \(r\), and any point \(P(x, y)\) on it (Fig. 1). Draw a horizontal line from \(C\) and a vertical line from \(P\): they meet at a right angle. The horizontal side is \(x - a\), the vertical side is \(y - b\) and the hypotenuse \(CP\) is the radius. Pythagoras gives \((x - a)^{2} + (y - b)^{2} = r^{2}\). Every point on the circle satisfies this, and no other point does, so it is the equation of the circle. With the centre at the origin it becomes \(x^{2} + y^{2} = r^{2}\).
Centre and radius: put them straight into \((x - a)^{2} + (y - b)^{2} = r^{2}\).
Centre and a point on the circle: \(r^{2}\) is the distance squared from the centre to the point.
Ends of a diameter: the centre is the MIDPOINT; the radius is half the length of the diameter.
Reading back: \((x + 3)^{2}\) is \((x - (-3))^{2}\), so \(a = -3\). The right-hand side is \(r^{2}\): if it is 16, then \(r = 4\).
Worked example From a centre and a radius
Write down the equation of the circle with centre \((3, -2)\) and radius 4, and give it in the form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).
The signs in the brackets are the OPPOSITE of the centre's coordinates: centre \((3, -2)\) gives \((x - 3)^{2} + (y + 2)^{2}\), not \((x + 3)^{2} + (y - 2)^{2}\). And the right-hand side is \(r^{2}\), not \(r\): radius 4 means \(= 16\), and \((x - 2)^{2} + y^{2} = 25\) has radius 5, not 25.
Section 02
From the general form back to the centre.
Expanding \((x - a)^{2} + (y - b)^{2} = r^{2}\) always gives the same shape: \(x^{2}\) and \(y^{2}\) with EQUAL coefficients, no \(xy\) term, some \(x\) and \(y\) terms and a constant. The List of formulas writes it as \(x^{2} + y^{2} + 2gx + 2fy + c = 0\). To find the centre and radius, either complete the square in \(x\) and in \(y\), or compare coefficients: the centre is \((-g, -f)\) and the radius is \(\sqrt{g^{2} + f^{2} - c}\). If the coefficients of \(x^{2}\) and \(y^{2}\) are not 1, divide the whole equation by them first.
Completing the square: \(x^{2} - 8x = (x - 4)^{2} - 16\); halve the coefficient of \(x\), square it and take it away.
Formula: \(2g\) is the coefficient of \(x\), so \(g\) is HALF of it; the centre is \((-g, -f)\).
Radius: \(r = \sqrt{g^{2} + f^{2} - c}\); note the MINUS sign on \(c\).
A real circle needs \(r^{2} > 0\), that is \(g^{2} + f^{2} - c > 0\). If \(r^{2} = 0\) the equation is a single point; if \(r^{2} < 0\) no point satisfies it.
Worked example Completing the square, then checking with the formula
Find the centre and radius of the circle \(x^{2} + y^{2} - 8x + 6y - 11 = 0\).
Find the set of values of \(k\) for which \(x^{2} + y^{2} - 4x + 10y + k = 0\) is the equation of a circle.
\(g = -2\), \(f = 5\), \(c = k\). A circle needs \(g^{2} + f^{2} - c > 0\): \(4 + 25 - k > 0\), so \(k < 29\).
Examiner trap
Two halving and sign slips cost marks here. The centre of \(x^{2} + y^{2} - 8x + 6y - 11 = 0\) is \((4, -3)\): not \((-8, 6)\) (forgetting to halve) and not \((-4, 3)\) (forgetting the minus signs in \((-g, -f)\)). And the radius uses \(-c\): with \(c = -11\) you ADD 11, giving \(\sqrt{36}\), not \(\sqrt{14}\). Forgetting to divide by 2 in \(2x^{2} + 2y^{2} + \dots\) gives the wrong centre and radius.
Stage 1 · Learn
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Find the centre and radius of the circle \((x + 2)^{2} + (y - 5)^{2} = 49\).
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Which is the equation of the circle with centre \((0, 3)\) and radius 4?
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Find the centre and radius of \(x^{2} + y^{2} + 10x - 4y + 4 = 0\).
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Where is the point \((3, 4)\) relative to the circle \(x^{2} + y^{2} = 20\)?
Section 03
Inside, on or outside, and the circle through three points.
To place a point, compare its distance squared from the centre, \(d^{2}\), with \(r^{2}\): \(d^{2} < r^{2}\) means inside, \(d^{2} = r^{2}\) on the circle, \(d^{2} > r^{2}\) outside. Equivalently, substitute the point into \(x^{2} + y^{2} + 2gx + 2fy + c\): a negative value means inside, zero means on, positive means outside. To find the circle through three points, use the fact that the perpendicular bisector of any chord passes through the centre (topic 7.3): find two perpendicular bisectors and solve them simultaneously.
Place a point: work out \((x - a)^{2} + (y - b)^{2}\) and compare it with \(r^{2}\).
Perpendicular bisector of a chord: through the midpoint, with gradient \(-\dfrac{1}{m}\) where \(m\) is the gradient of the chord.
Three points: two perpendicular bisectors meet at the centre; the radius is the distance from the centre to any of the points.
Angle in a semicircle: if \(AB\) is a diameter and \(C\) is on the circle, angle \(ACB = 90^{\circ}\). So if the gradients of \(AC\) and \(BC\) multiply to \(-1\), \(AB\) is a diameter. Use this as a check, or as a short cut when the triangle has a right angle.
Worked example Inside, on or outside?
For the circle \((x - 2)^{2} + (y + 1)^{2} = 25\), decide whether each point is inside, on or outside: \(A(5, 3)\), \(B(6, 1)\), \(C(-2, 3)\).
\(r^{2} = 25\).
\(A\): \((5 - 2)^{2} + (3 + 1)^{2} = 9 + 16 = 25\), equal to \(r^{2}\), so on the circle.
Decide whether the point \((1, 6)\) lies inside, on or outside the circle \(x^{2} + y^{2} - 4x + 2y - 20 = 0\).
Substitute: \(1 + 36 - 4 + 12 - 20 = 25 > 0\), so outside. (In centre-radius form the circle is \((x - 2)^{2} + (y + 1)^{2} = 25\), and \(d^{2} = 1 + 49 = 50 > 25\).)
Examiner trap
Compare like with like: \(d^{2}\) with \(r^{2}\), or \(d\) with \(r\), never \(d^{2}\) with \(r\). For \(x^{2} + y^{2} = 20\) and the point \((3, 4)\), \(d^{2} = 25 > 20\), so it is outside; comparing \(d = 5\) with 20 wrongly says inside. In the three-point method the perpendicular bisector must go through the MIDPOINT of the chord, not through one of its ends.
Stage 2 · Exam
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The points \((-3, 2)\) and \((5, 8)\) are the ends of a diameter of a circle. Find the equation of the circle.
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Find the set of values of \(k\) for which \(x^{2} + y^{2} - 6x + 8y + k = 0\) represents a circle.
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A circle has centre \((2, -3)\) and passes through \((5, 1)\). Find its equation in the form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\).
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A circle passes through \(O(0, 0)\), \(A(8, 0)\) and \(B(0, 6)\). Find its equation.
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