Mg10 - Circle equations, centres and radii

Mg10 - Circle equations, centres and radii

Connect the fixed-distance meaning of a circle to its algebraic equation. You will complete squares to determine centres and radii, construct circles from points or diameters, and draw their important features accurately.

A circle as a distance locus

Can x2+y2+4x−6y−12=0x^2+y^2+4x-6y-12=0 describe a circle even though neither its centre nor its radius is obvious? Start with a clearer description: all points 44 units from (2,−3)(2,-3). A point directly to its right at (6,−3)(6,-3) is one of them.

You already know

For two coordinate points, squared distance is the sum of the squared horizontal and vertical changes. Completing the square rewrites x2+4xx^2+4x as (x+2)2−4(x+2)^2-4, without changing its value.

The centre stays fixed; the point on the circumference changes position.

Circle

A circle is the locus of points at a fixed positive distance rr from a fixed point (a,b)(a,b), the centre. The distance rr is its radius. The centre is not itself on the circumference.

Distance gives (x−a)2+(y−b)2=r\sqrt{(x-a)^2+(y-b)^2}=r. Squaring is reversible because the two sides are non-negative.

Circle equation

For real a,ba,b and r>0r>0,

(x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2

The centre is (a,b)(a,b). At the origin, the form reduces to x2+y2=r2x^2+y^2=r^2.

Not in the formula sheet: learn it.

An equation is also a test: substitute a point to find whether its squared distance equals the squared radius.

Worked example 1: verifying a point

Question. Write the circle equation for centre (2,−3)(2,-3), radius 44, and verify that (6,−3)(6,-3) lies on it.

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

(x−2)2+(y+3)2=16use r2=42(6−2)2+(−3+3)2=16substitute the point\begin{aligned} (x-2)^2+(y+3)^2&=16 && \text{use }r^2=4^2 \\ (6-2)^2+(-3+3)^2&=16 && \text{substitute the point} \end{aligned}

Why is the second bracket y+3y+3? The centre's vertical coordinate is −3-3, so you subtract −3-3.

Answer. (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16; the substitution verifies that (6,−3)(6,-3) lies on it.

Check. The point is 44 units horizontally from the centre, agreeing with the radius.

Change only the radius from 44 to 55, keeping the centre and test point unchanged.

Worked example 2: comparing a distance with a larger radius

Question. Write the equation for centre (2,−3)(2,-3), radius 55, and determine the position of (6,−3)(6,-3).

Plan. Compare the same squared distance with the new squared radius.

(x−2)2+(y+3)2=25change the radius(6−2)2+(−3+3)2=16<25compare the squares\begin{aligned} (x-2)^2+(y+3)^2&=25 && \text{change the radius} \\ (6-2)^2+(-3+3)^2&=16<25 && \text{compare the squares} \end{aligned}

Why does the inequality establish the position? Distances are non-negative, so a smaller square means a smaller distance.

Answer. The equation is (x−2)2+(y+3)2=25(x-2)^2+(y+3)^2=25, and the point is inside.

Check. Its distance is 44, less than 55.

More generally, a squared distance smaller than r2r^2 puts a point inside, and a larger one puts it outside. Equality gives the circumference. These comparisons are consequences of the distance definition, rather than separate supplied formulas.

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 put 42=164^2=16 on the right.

Use a different centre and a point whose two coordinate changes are nonzero.

Recovering the centre by completing squares

Completing squares changes the appearance of an equation, rather than its points. Group the xx terms and yy terms separately; balance every constant you introduce.

Expanded circle equations

Completing the squares in x2+y2+Dx+Ey+F=0x^2+y^2+Dx+Ey+F=0 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

A circle requires a positive right side. Zero describes one point; a negative value describes no real points.

Not in the formula sheet: learn it.

The signs in the brackets reveal the centre. The whole right side is the squared radius, so take its positive square root.

Worked example 3: a hidden centre and radius

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

Plan. Complete each square, then identify centre and positive radius.

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

Why is the centre (−2,3)(-2,3)? Those coordinates make both squared brackets zero.

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

Check. Expanding the completed form gives x2+y2+4x−6y+13=25x^2+y^2+4x-6y+13=25, which returns the given equation.

Change only the equation's common scale by multiplying every term by 33. Its solution set remains unchanged.

Worked example 4: normalising the squared terms

Question. Determine the centre and radius of 3x2+3y2+12x−18y−36=03x^2+3y^2+12x-18y-36=0.

Plan. Divide all terms by 33 before completing squares.

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

Why divide every term? Dividing only the squared terms would produce a different equation and a different circle.

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

Check. Multiply the expanded circle equation by 33; all five coefficients match the question.

For example, the different equation 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. Unequal coefficients on the two squares cannot be read directly as the circle form.

A small constant can make a big geometric difference. In x2+y2+4x−6y+F=0x^2+y^2+4x-6y+F=0, the squared radius is 13−F13-F. Changing FF to 1313 leaves just (−2,3)(-2,3); changing it to 1414 leaves no real points. A sum of real squares cannot be negative. The same test gives no real circle for x2+y2+4x−10y+30=0x^2+y^2+4x-10y+30=0, whose completed right side is −1-1.

Watch out

Wrong: use (2,−3)(2,-3) as the centre of (x+2)2+(y−3)2=25(x+2)^2+(y-3)^2=25. Right: the centre is (−2,3)(-2,3), the coordinates that make both brackets vanish.

Show the completed form so that both the signs and the radius can be checked.

Constructing and drawing a circle

A circle equation needs its centre and squared radius. A known centre and one circumference point provide them; a diameter provides both through its midpoint and length.

Worked example 5: a radius from coordinates

Question. Determine the circle with centre (−1,2)(-1,2) passing through (2,6)(2,6).

Plan. Find the squared centre-to-point distance, then write the equation.

r2=(2+1)2+(6−2)2distance from the centre=9+16=25square and add(x+1)2+(y−2)2=25use centre–radius form\begin{aligned} r^2&=(2+1)^2+(6-2)^2 && \text{distance from the centre} \\ &=9+16=25 && \text{square and add} \\ (x+1)^2+(y-2)^2&=25 && \text{use centre–radius form} \end{aligned}

Why keep the squared distance? It is exactly the number needed on the right side.

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

Check. The displacement (3,4)(3,4) has length 55, so the result has the required radius.

Change only the form of the geometric information: give the opposite diameter endpoint in place of the centre.

Worked example 6: recovering the centre from a diameter

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

Plan. Use their midpoint as centre and measure from it to one endpoint.

C=(−4+22,−2+62)=(−1,2)diameter midpointr2=(2+1)2+(6−2)2=25squared radius(x+1)2+(y−2)2=25form the equation\begin{aligned} C&=\left(\frac{-4+2}{2},\frac{-2+6}{2}\right)=(-1,2) && \text{diameter midpoint} \\ r^2&=(2+1)^2+(6-2)^2=25 && \text{squared radius} \\ (x+1)^2+(y-2)^2&=25 && \text{form the equation} \end{aligned}

Why use half the diameter? The two endpoints lie on opposite sides of the centre, each one radius away.

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

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

Another diameter, from (−3,1)(-3,1) to (5,7)(5,7), gives centre (1,4)(1,4) and equation (x−1)2+(y−4)2=25(x-1)^2+(y-4)^2=25. One circumference point alone would allow many different centres and circles.

[DIAGRAM: asset_name: A circle recovered from completed squares; caption: Equal axis scales show the centre and radius without distortion; asset_slug: h640-centre-radius; description: Equal-scale axes show the circle centred at minus 2 comma 3 with radius 5. Its four extreme points are minus 7 comma 3, 3 comma 3, minus 2 comma 8 and minus 2 comma minus 2. A horizontal radius and the centre are labelled.]
Equal axis scales show the centre and radius without distortion

To draw the opening circle, first mark (−2,3)(-2,3), then its four extremes (−7,3)(-7,3), (3,3)(3,3), (−2,8)(-2,8), (−2,−2)(-2,-2). Use the same physical scale on both axes and draw a smooth circumference; unequal scales would display an ellipse. Use a compass for an accurate drawing when possible.

Intersections with axes come from substitutions. Putting y=0y=0 gives (x+2)2=16(x+2)^2=16, so x=−6,2x=-6,2. Putting x=0x=0 gives (y−3)2=21(y-3)^2=21, so y=3±21y=3\pm\sqrt{21}. A circle may miss an axis, so do not assume every sketch needs two intercepts on each. These construction and intercept methods follow from the equation and are not printed as formulas in the sheet.

Exam tip

When MEI asks you to determine a diameter-defined circle, show how you found the midpoint and radius. In a sketch, label the centre and essential features; for a drawing, choose a scale that preserves the circular shape.

Use a different diameter and finish with the full equation.

Quick Recap

  • The circle equation fixes a positive distance from its centre; its right side is the squared radius.
  • Complete both squares to recover the centre with the correct signs.
  • A zero right side leaves one point; a negative one leaves no real points.
  • Diameter endpoints give a midpoint centre and half-distance radius.
  • Preserve equal axis scales and find intercepts by substitution.

The opening equation is (x+2)2+(y−3)2=25(x+2)^2+(y-3)^2=25, so it describes a circle with centre (−2,3)(-2,3) and radius 55.