CG3 - The equation of a circle

CG3 - The equation of a circle

A circle equation is a distance statement in disguise. You will write one from a centre, a point or a diameter, uncover the centre by completing the square, and find where a line meets a circle. Keeping squared lengths exact makes the geometry much simpler than measuring a drawing.

Centre, radius and equation

Can one equation describe every point exactly 55 units from (1,−2)(1, -2), without accidentally including the points inside? Try P(4,2)P(4, 2): its horizontal and vertical separations from (1,−2)(1, -2) are 33 and 44, so its distance is 55. The equation needs to make that same test for any point (x,y)(x, y).

You already know

Squared distance between (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is (x2−x1)2+(y2−y1)2(x_2-x_1)^2+(y_2-y_1)^2. A diameter's centre is its mid-point, and its radius is half its length.

Replace the particular point (4,2)(4, 2) by (x,y)(x, y). Requiring squared distance 2525 gives (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25. Points inside give a smaller left-hand side; points outside give a larger one. Equality picks out the boundary alone.

Circle

A circle is the set of points in a plane at a fixed positive distance, the radius, from a fixed point, the centre. The filled disc includes the inside; the circle itself does not.

Pythagoras supplies the equation for any centre. A surprising detail is that the right-hand side needs only the squared radius, so you can write an exact circle equation without ever calculating the square root of the radius squared.

Equation of a circle

(x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2

The centre is (a,b)(a, b), and the radius is r>0r>0. A point lies on the circle exactly when its coordinates satisfy the equation.
Not on the formulae sheet: learn it.

The signs inside the brackets express differences from the centre. A centre below the horizontal axis therefore produces a plus sign in the second bracket.

[DIAGRAM: asset_name: A circle from a radius or a diameter; asset_slug: cg19-circle-centre; caption: The same circle can be defined by its centre and a point or by the endpoints of a diameter; description: The circle has centre C at 1 comma minus 2 and radius 5. Diameter endpoints A at minus 2 comma 2 and B at 4 comma minus 6 lie on opposite sides of C. Another radius joins C to P at 4 comma 2, with perpendicular horizontal and vertical changes 3 and 4.]
The same circle can be defined by its centre and a point or by the endpoints of a diameter

Worked example 1: a centre and a point

Question. A circle has centre C(1,−2)C(1, -2) and passes through P(4,2)P(4, 2). Determine its equation.

Plan. Find the squared distance CP2CP^2, then place the centre and squared radius in the circle equation.

r2=(4−1)2+(2−(−2))2squared radius=32+42=25evaluate exactly(x−1)2+(y+2)2=25centre (1,−2)\begin{aligned} r^2&=(4-1)^2+(2-(-2))^2 && \text{squared radius}\\ &=3^2+4^2=25 && \text{evaluate exactly}\\ (x-1)^2+(y+2)^2&=25 && \text{centre }(1, -2) \end{aligned}

Why stop at r2r^2? The equation needs 2525, not an unnecessary calculation of rr.

Answer. (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25.

Check. At PP, (4−1)2+(2+2)2=25(4-1)^2+(2+2)^2=25, so PP is on the circle. At the centre, the left-hand side is 00, so the centre is not on it.

The next example changes how the circle is given: the endpoints of a diameter replace the centre and one point. The diagram shows why the mid-point must come first.

Worked example 2: a diameter determines the circle

Question. A(−2,2)A(-2, 2) and B(4,−6)B(4, -6) are the ends of a diameter. Determine the circle's equation.

Plan. Find the centre as the mid-point, then use one endpoint to find the squared radius.

C=(−2+42,2−62)=(1,−2)mid-point of diameterr2=(−2−1)2+(2+2)2=25centre to endpoint(x−1)2+(y+2)2=25insert centre and r2\begin{aligned} C&=\left(\frac{-2+4}{2}, \frac{2-6}{2}\right)=(1, -2) && \text{mid-point of diameter}\\ r^2&=(-2-1)^2+(2+2)^2=25 && \text{centre to endpoint}\\ (x-1)^2+(y+2)^2&=25 && \text{insert centre and }r^2 \end{aligned}

Why not use AB2AB^2 for r2r^2? ABAB is the diameter, so its squared length is four times the squared radius.

Answer. (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25.

Check. AB2=62+(−8)2=100AB^2=6^2+(-8)^2=100, giving r2=100/4=25r^2=100/4=25, as required.

Reading a sign as a coordinate is an easy way to put the right circle in the wrong place.

Watch out

Wrong: (x+3)2+(y−4)2=16(x+3)^2+(y-4)^2=16 has centre (3,−4)(3, -4) and radius 1616. Right: its centre is (−3,4)(-3, 4) and radius is 44. The brackets vanish at the centre, and the right-hand side is the radius squared.

Try a new diameter with new coordinates. You can still keep the equation tidy.

Uncovering the centre by completing the square

An expanded equation hides the centre in its linear terms; completing the square brings the geometry back into view.

You already know

Completing the square gives x2+px=(x+p2)2−p24x^2+px=(x+\frac p2)^2-\frac{p^2}{4}. Apply it separately to the xx terms and the yy terms. Keep both corrections.

For example, x2−6x=(x−3)2−9x^2-6x=(x-3)^2-9 and y2+4y=(y+2)2−4y^2+4y=(y+2)^2-4. The corrections total −13-13, which must be included when the constant is moved to the other side. The completing-square result is not on the formulae sheet: learn it.

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