P2 3.1 - Equations of circles

P2 3.1 - Equations of circles

Connect a circle's equation to its centre and radius, complete squares to reveal that information, and determine equations from given points. Keeping squared distances visible prevents unnecessary square roots and helps you check the geometry.

Centre-radius form

How can one equation describe every point exactly 55 units from (2,1)(2,1)? For the point (5,5)(5,5), the horizontal and vertical differences from (2,1)(2,1) are 33 and 44. Pythagoras gives 32+42=523^2+4^2=5^2.

Replace that particular point by a moving point (x,y)(x,y). Its squared distance from (2,1)(2,1) is (x−2)2+(y−1)2(x-2)^2+(y-1)^2, so setting this equal to 2525 describes the whole circle.

[DIAGRAM: asset_name: A circle as a fixed distance; asset_slug: circle-fixed-distance; caption: The differences from the centre form a right triangle whose hypotenuse is a radius.; description: Equal-scale coordinate diagram showing centre C at (2,1), point P at (5,5), a radius of length 5, and horizontal and vertical components 3 and 4. The circle has equation (x-2) squared plus (y-1) squared equals 25.]
The differences from the centre form a right triangle whose hypotenuse is a radius.

Centre-radius equation

A circle with centre (a,b)(a,b) and radius r>0r>0 has equation

(x−a)2+(y−b)2=r2.(x-a)^2+(y-b)^2=r^2.

A point lies on the circle exactly when its coordinates satisfy this equation. A smaller squared distance places it inside; a larger one places it outside.

Not in the formula booklet: learn it.

The signs in the brackets express differences from the centre, not the centre's coordinates directly.

Worked example 1: writing an equation

Question. Find the equation of the circle with centre (2,1)(2,1) and radius 55.

Plan. Subtract the centre coordinates from x,yx,y and equate the sum of squares to 525^2.

(x−2)2+(y−1)2=25.(x-2)^2+(y-1)^2=25.

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

Check. The rightmost point is (7,1)(7,1). Substitution gives (7−2)2+(1−1)2=25(7-2)^2+(1-1)^2=25, and its distance from the centre is 55.

Change only the sign of the centre's first coordinate. The horizontal bracket changes from subtraction to addition.

Worked example 2: a negative centre coordinate

Question. Find the equation of the circle with centre (−2,1)(-2,1) and radius 55.

Plan. Use x−(−2)=x+2x-(-2)=x+2, retaining the same vertical difference and radius.

(x+2)2+(y−1)2=25.(x+2)^2+(y-1)^2=25.

Answer. (x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25. Reading this equation backwards gives centre (−2,1)(-2,1) and radius 25=5\sqrt{25}=5.

Check. Its rightmost point is (3,1)(3,1). Substitution gives 52+02=255^2+0^2=25; the horizontal distance from −2-2 to 33 is 55.

A circle's centre is usually not on its circumference: substituting the centre makes the left-hand side zero.

Watch out

In (x+2)2+(y−1)2=25(x+2)^2+(y-1)^2=25, the centre is (−2,1)(-2,1), not (2,−1)(2,-1), and the radius is 55, not 2525. The right-hand side is the radius squared.

Read both pieces of information from this new equation.

Completing squares

An expanded circle equation hides its centre, but completing one square in xx and one in yy reveals it.

Recall x2−6x=(x−3)2−9x^2-6x=(x-3)^2-9. The subtracted 99 compensates for the extra 99 introduced by expanding the bracket. Complete both squares before collecting constants.

Expanded circle equation

For

x2+y2+Dx+Ey+F=0,x^2+y^2+Dx+Ey+F=0,

completion of squares gives

(x+D2)2+(y+E2)2=D2+E24−F.\left(x+\frac D2\right)^2+\left(y+\frac E2\right)^2 =\frac{D^2+E^2}{4}-F.

The centre is (−D/2,−E/2)(-D/2,-E/2). A positive right-hand side gives a circle, zero gives a single point, and a negative value gives no real points.

Not in the formula booklet: learn it.

The equal coefficients of x2x^2 and y2y^2 matter. If they share a common nonzero coefficient, divide it out first. An equation such as x2+2y2=1x^2+2y^2=1 is not a circle.

Worked example 3: revealing centre and radius

Question. Find the centre and radius of x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0.

Plan. Group by variable and complete both squares.

x2−6x+y2+4y=12(x−3)2−9+(y+2)2−4=12(x−3)2+(y+2)2=25\begin{aligned} x^2-6x+y^2+4y&=12\\ (x-3)^2-9+(y+2)^2-4&=12\\ (x-3)^2+(y+2)^2&=25 \end{aligned}

Answer. Centre (3,−2)(3,-2), radius 55.

Check. Expanding the final equation gives x2−6x+9+y2+4y+4=25x^2-6x+9+y^2+4y+4=25, or x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0.

Change only the original constant from −12-12 to 44. The linear terms, and therefore the centre, stay the same; the radius changes.

Worked example 4: changing the constant

Question. Find the centre and radius of x2+y2−6x+4y+4=0x^2+y^2-6x+4y+4=0.

Plan. Complete the same squares, but keep the new constant on the right.

x2−6x+y2+4y=−4(x−3)2−9+(y+2)2−4=−4(x−3)2+(y+2)2=9\begin{aligned} x^2-6x+y^2+4y&=-4\\ (x-3)^2-9+(y+2)^2-4&=-4\\ (x-3)^2+(y+2)^2&=9 \end{aligned}

Answer. Centre (3,−2)(3,-2), radius 33.

Check. The point (6,−2)(6,-2) is 33 units from the centre. The original left-hand side is 36+4−36−8+4=036+4-36-8+4=0, as required.

The constant controls the size only after the square-completion adjustments have been included.

Watch out

The line x2−6x=(x−3)2x^2-6x=(x-3)^2 is wrong: the right side contains an extra 99. Use (x−3)2−9(x-3)^2-9. A missed compensation changes the radius.

Complete both squares before taking a square root.

Circle through given points

A known centre and one point determine the radius; the endpoints of a known diameter determine both the centre and the radius.

If the centre is (a,b)(a,b) and (u,v)(u,v) lies on the circle, then r2=(u−a)2+(v−b)2r^2=(u-a)^2+(v-b)^2. If A,BA,B are opposite ends of a diameter, the centre is their midpoint. Either use the distance from that midpoint to an endpoint, or divide the squared diameter by 44.

Diameter endpoints

For diameter endpoints A(x1,y1)A(x_1,y_1), B(x2,y2)B(x_2,y_2),

C=(x1+x22,y1+y22),r2=(x2−x1)2+(y2−y1)24.C=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\qquad r^2=\frac{(x_2-x_1)^2+(y_2-y_1)^2}{4}.

This uses the midpoint rule and diameter=2r\text{diameter}=2r.

Not in the formula booklet: learn it.

Two points stated only to lie on a circle do not determine a unique circle. You must use the additional information, such as “diameter” or a condition on its centre.

Worked example 5: diameter information

Question. A(−1,2)A(-1,2) and B(5,10)B(5,10) are the endpoints of a diameter. Find the circle's equation.

Plan. Find the midpoint, then find the squared radius from the centre to AA.

C=(−1+52,2+102)=(2,6)r2=(−1−2)2+(2−6)2=9+16=25\begin{aligned} C&=\left(\frac{-1+5}{2},\frac{2+10}{2}\right)=(2,6)\\ r^2&=(-1-2)^2+(2-6)^2=9+16=25 \end{aligned}

Answer. (x−2)2+(y−6)2=25(x-2)^2+(y-6)^2=25.

Check. The squared diameter is 62+82=100=4r26^2+8^2=100=4r^2. Substituting either endpoint into the circle equation gives 2525.

Move only BB's vertical coordinate from 1010 to 66. Both the midpoint and radius must be recalculated.

Worked example 6: an exact radius and an extreme point

Question. A(−1,2)A(-1,2) and B(5,6)B(5,6) are the endpoints of a diameter. Find the circle's equation and the exact point on it nearest the xx-axis.

Plan. Find its centre and squared radius, then look at the lowest point of the circle.

C=(2,4),r2=(−1−2)2+(2−4)2=13.C=(2,4),\qquad r^2=(-1-2)^2+(2-4)^2=13.

Its lowest height is 4−13>04-\sqrt{13}>0, so the whole circle lies above the xx-axis. The nearest point is directly below the centre, one radius away.

Answer. The equation is (x−2)2+(y−4)2=13(x-2)^2+(y-4)^2=13, and the nearest point is (2,4−13)(2,4-\sqrt{13}).

Check. At that point, the squared differences are 00 and 1313, so it lies on the circle. The other vertical extreme, (2,4+13)(2,4+\sqrt{13}), is farther from the axis.

“Lowest point” and “nearest the xx-axis” coincide here because the circle is wholly above that axis. A circle crossing the axis has distance zero at its intersection points instead.

Watch out

The distance between diameter endpoints is 2r2r. If you calculate the squared distance between them, divide by 44, not by 22, to obtain r2r^2.

Use the stated diameter rather than treating the endpoints as a centre and a point.

Quick Recap

  • A circle is a set of points at a fixed distance from its centre.
  • Read its centre and squared radius from (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2.
  • Complete both squares to reveal an expanded equation's geometry.
  • A diameter's midpoint is the centre; its length is twice the radius.

The opening set of points is exactly the circle (x−2)2+(y−1)2=25(x-2)^2+(y-1)^2=25.