1.03d, 1.03e - Circle equations, centres and radii

1.03d, 1.03e - Circle equations, centres and radii

Read a circle equation as a statement about distance. You will construct equations from geometric information, uncover centres and radii by completing squares, and draw circles with the correct position, size and scale.

A circle as a distance locus

Where is the centre of x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0, and why will its radius be 55, rather than 2525? Start with a circle whose geometry is visible: every point 44 units from (−2,3)(-2,3) belongs to one circumference.

You already know

The distance from (a,b)(a,b) to (x,y)(x,y) is (x−a)2+(y−b)2\sqrt{(x-a)^2+(y-b)^2}. Complete a square by adding and subtracting the same number, for example x2−6x=(x−3)2−9x^2-6x=(x-3)^2-9.

The centre is fixed, while a point on the circumference moves.

Circle

A circle is the set, or locus, of points at a fixed positive distance rr from a fixed point CC, its centre. The distance rr is its radius. The centre itself is inside the circle, rather than on its circumference.

The distance formula gives (x−a)2+(y−b)2=r\sqrt{(x-a)^2+(y-b)^2}=r. Both sides are non-negative, so squaring is reversible.

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

For centre (0,0)(0,0), this becomes x2+y2=r2x^2+y^2=r^2.

Not in the formula sheet: learn it.

Substitution tests a point's distance without requiring a sketch.

Worked example 1: a point on a circle

Question. Write the equation of the circle with centre (−2,3)(-2,3), radius 44, and decide whether (2,3)(2,3) lies on it.

Plan. Substitute the centre and squared radius, then calculate the point's squared distance.

(x+2)2+(y−3)2=16use r2=42(2+2)2+(3−3)2=16test the point\begin{aligned} (x+2)^2+(y-3)^2&=16 && \text{use }r^2=4^2 \\ (2+2)^2+(3-3)^2&=16 && \text{test the point} \end{aligned}

Why is the first bracket x+2x+2? You subtract the centre's coordinate −2-2.

Answer. (x+2)2+(y−3)2=16(x+2)^2+(y-3)^2=16; the point is on the circumference.

Check. The point is exactly 44 units horizontally from the centre.

Change only the radius, from 44 to 55. Keep the centre and test point fixed.

Worked example 2: the same point is now inside

Question. Write the equation with centre (−2,3)(-2,3), radius 55, and classify (2,3)(2,3).

Plan. Change the squared radius and compare the same squared distance with it.

(x+2)2+(y−3)2=25new squared radius(2+2)2+(3−3)2=16<25compare squared distances\begin{aligned} (x+2)^2+(y-3)^2&=25 && \text{new squared radius} \\ (2+2)^2+(3-3)^2&=16<25 && \text{compare squared distances} \end{aligned}

Why is comparison of squares enough? Both the distance and radius are non-negative, so the ordering is preserved.

Answer. (x+2)2+(y−3)2=25(x+2)^2+(y-3)^2=25; (2,3)(2,3) is inside.

Check. Its distance is 44, less than the new radius 55.

In general, a squared distance less than r2r^2 is inside, equality is on the circumference, and greater than r2r^2 is outside. These comparisons follow directly from distance; they are not extra formula-sheet results.

Watch out

Wrong: centre (−2,3)(-2,3), radius 44, gives (x−2)2+(y+3)2=4(x-2)^2+(y+3)^2=4. Right: subtract the centre coordinates and square the radius, giving (x+2)2+(y−3)2=16(x+2)^2+(y-3)^2=16.

Use a new centre, then give a calculation for the point decision.

Recovering the centre by completing squares

An expanded equation hides two completed squares. Recover them separately for xx and yy, then collect the constants on the right.

Completing the two squares

For real constants D,E,FD,E,F,

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

can be written

(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

A positive right side gives a circle; zero gives one point; a negative value gives no real points.

Not in the formula sheet: learn it.

You need not memorise the expanded constant expression if you can complete the squares reliably. A radius is the square root of the resulting positive number.

Worked example 3: uncovering the opening circle

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

Plan. Complete each square and balance the constants.

(x−3)2−9+(y+2)2−4−12=0complete both squares(x−3)2+(y+2)2=25collect constantsr=25=5take the positive root\begin{aligned} (x-3)^2-9+(y+2)^2-4-12&=0 && \text{complete both squares} \\ (x-3)^2+(y+2)^2&=25 && \text{collect constants} \\ r&=\sqrt{25}=5 && \text{take the positive root} \end{aligned}

Why is the centre (3,−2)(3,-2)? The brackets subtract those coordinates: y+2=y−(−2)y+2=y-(-2).

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

Check. Expanding gives x2−6x+9+y2+4y+4=25x^2-6x+9+y^2+4y+4=25, hence the given equation.

Change only the common scale of the equation: multiply every term by 22. The geometric set must stay the same.

Worked example 4: a common squared-term coefficient

Question. Find the centre and radius of 2x2+2y2−12x+8y−24=02x^2+2y^2-12x+8y-24=0.

Plan. Divide the entire equation by 22, then complete the squares.

x2+y2−6x+4y−12=0divide every term by 2(x−3)2+(y+2)2=25complete the squares\begin{aligned} x^2+y^2-6x+4y-12&=0 && \text{divide every term by }2 \\ (x-3)^2+(y+2)^2&=25 && \text{complete the squares} \end{aligned}

Why divide first? The standard circle form has coefficient 11 on each squared variable.

Answer. Centre (3,−2)(3,-2), radius 55, unchanged by the nonzero common factor.

Check. Multiplying the completed equation by 22 and expanding returns the question's equation.

For another common-factor example, 2x2+2y2−8x+12y−6=02x^2+2y^2-8x+12y-6=0 gives (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16, centre (2,−3)(2,-3), radius 44. Divide every term, rather than only the squared terms. Unequal squared coefficients cannot be read directly as this circle form.

There is a striking boundary case: changing only a constant can shrink a whole circle to one point or erase it. In x2+y2−6x+4y+F=0x^2+y^2-6x+4y+F=0, the completed right side is 13−F13-F. At F=13F=13, only (3,−2)(3,-2) remains; at F=14F=14, there are no real points. Similarly x2+y2+4x−10y+30=0x^2+y^2+4x-10y+30=0 gives (x+2)2+(y−5)2=−1(x+2)^2+(y-5)^2=-1, so it is not a real circle.

Watch out

Wrong: the right side 2525 is the radius. Right: it is r2r^2, so the radius is 55. A negative right side has no real square-root radius.

Complete both squares before reading off the centre and radius.

Constructing and drawing a circle

To construct a circle equation, identify a centre and one radius. A point on the circumference gives the radius if the centre is known; diameter endpoints give the centre as well.

Worked example 5: centre and one point

Question. Find the circle with centre (1,−2)(1,-2) passing through (4,2)(4,2).

Plan. Calculate the squared centre-to-point distance and use it as r2r^2.

r2=(4−1)2+(2+2)2distance from the centre=32+42=25square and add(x−1)2+(y+2)2=25use centre–radius form\begin{aligned} r^2&=(4-1)^2+(2+2)^2 && \text{distance from the centre} \\ &=3^2+4^2=25 && \text{square and add} \\ (x-1)^2+(y+2)^2&=25 && \text{use centre–radius form} \end{aligned}

Why calculate r2r^2 directly? The circle equation needs that value, so a square root and later squaring are unnecessary.

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

Check. Substitution of (4,2)(4,2) gives 9+16=259+16=25.

Change only how the centre is supplied: replace it by the opposite diameter endpoint, retaining (4,2)(4,2) on the circle.

Worked example 6: endpoints of a diameter

Question. Find the circle with diameter endpoints (−2,−6)(-2,-6) and (4,2)(4,2).

Plan. Find the diameter's midpoint, then its squared radius from one endpoint.

C=(−2+42,−6+22)=(1,−2)midpoint of diameterr2=(4−1)2+(2+2)2=25centre-to-endpoint distance(x−1)2+(y+2)2=25form the equation\begin{aligned} C&=\left(\frac{-2+4}{2},\frac{-6+2}{2}\right)=(1,-2) && \text{midpoint of diameter} \\ r^2&=(4-1)^2+(2+2)^2=25 && \text{centre-to-endpoint distance} \\ (x-1)^2+(y+2)^2&=25 && \text{form the equation} \end{aligned}

Why use the midpoint? A diameter passes through the centre and has its endpoints one radius away on opposite sides.

Answer. (x−1)2+(y+2)2=25(x-1)^2+(y+2)^2=25, the same circle as before.

Check. The diameter length is 62+82=10\sqrt{6^2+8^2}=10, twice the radius 55.

For diameter endpoints (−3,1)(-3,1), (5,7)(5,7), the same construction gives centre (1,4)(1,4), diameter 1010 and equation (x−1)2+(y−4)2=25(x-1)^2+(y-4)^2=25. A single point without a centre or other restrictions would not determine one circle.

[DIAGRAM: asset_name: A circle recovered from completed squares; caption: Equal scales preserve the circle's shape and show its centre and radius; asset_slug: h240-centre-radius; description: Equal-scale axes show the circle centred at 3 comma minus 2 with radius 5. The four extreme points are minus 2 comma minus 2, 8 comma minus 2, 3 comma 3, and 3 comma minus 7. A horizontal radius and the centre are labelled.]
Equal scales preserve the circle's shape and show its centre and radius

To draw the opening circle, mark its centre (3,−2)(3,-2) and points one radius left, right, up and down: (−2,−2)(-2,-2), (8,−2)(8,-2), (3,3)(3,3), (3,−7)(3,-7). Use equal physical scales on the axes, then a smooth circle through those points. Unequal scales make a circle appear elliptical. For an accurate drawing, use a compass where the scale permits.

Find intercepts by substitution rather than guessing them from a sketch. Setting x=0x=0 gives (y+2)2=16(y+2)^2=16, so y=2,−6y=2,-6. Setting y=0y=0 gives (x−3)2=21(x-3)^2=21, so x=3±21x=3\pm\sqrt{21}. Some circles do not cross an axis at all. These constructions and intercept calculations are not in the formula sheet; derive them from the equation.

Watch out

Wrong: use the whole endpoint distance as the radius of a diameter-defined circle. Right: halve the diameter, or calculate the distance from its midpoint to an endpoint.

Use a fresh diameter. Give a complete equation rather than only the centre and radius.

Quick Recap

  • A circle is a fixed positive distance from its centre; square that radius in its equation.
  • Complete the xx and yy squares separately to reveal the centre and squared radius.
  • A zero squared-radius value gives one point; a negative value gives no real points.
  • Diameter endpoints give a midpoint centre and a radius equal to half their distance.
  • Draw with equal axis scales and calculate intercepts by setting one coordinate to zero.

The opening equation becomes (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25: its centre is (3,−2)(3,-2) and its radius is 25=5\sqrt{25}=5.