A16 - Equation of a circle and tangents (Higher)

A16 - Equation of a circle and tangents (Higher)

A circle centred at the origin has a short equation built from Pythagoras. You will recognise that equation, check whether points lie on the circle, and find tangent equations using perpendicular gradients. The coordinate rule links a curved shape to the straight line that just touches it.

A circle centred at the origin

The point (3,4)(3,4) lies on a circle of radius 55 centred at the origin. What equation describes the straight line that just touches the circle there? A line through that point could cross the circle, so we need a way to pick the tangent.

You already know

Pythagoras gives the distance between two points from their horizontal and vertical separation. A circle is the set of points at a fixed distance, its radius, from its centre. A tangent is perpendicular to the radius at its point of contact.

Let P=(x,y)P=(x,y) lie on a circle of radius rr centred at O=(0,0)O=(0,0). The horizontal and vertical distances from OO are ∣x∣|x| and ∣y∣|y|. Pythagoras gives x2+y2=r2x^2+y^2=r^2; squaring removes the signs. If a coordinate is zero, the same equation still holds on an axis.

[DIAGRAM: asset_name: Pythagoras gives the circle equation; asset_slug: aqa8300_g27_equation; caption: Every point on the circle has the same distance from the origin; description: Equal-scale coordinate axes and the circle x squared plus y squared equals 25. A radius from O to P at 3,4 is labelled 5. Dashed perpendiculars show horizontal distance 3 and vertical distance 4, with a right-angle square at 3,0. Axis intercepts plus and minus5 are labelled.]
Every point on the circle has the same distance from the origin

This is an equation for the circumference. Points strictly inside have a smaller sum of coordinate squares; points outside have a larger sum.

Equation of a circle

For a circle with centre (0,0)(0,0) and radius r>0r>0,

x2+y2=r2x^2+y^2=r^2

The right-hand side is the radius squared. The intercepts are (r,0),(−r,0),(0,r),(0,−r)(r,0),(-r,0),(0,r),(0,-r).

Not on the formulae sheet: learn it. It follows from Pythagoras, which is on the 2027 formulae sheet.

An equal scale on both axes is essential when sketching a circle; unequal scales would make its plotted shape look stretched.

Worked example 1: writing the equation

Question. Non-calculator. A circle has centre (0,0)(0,0) and radius 66. Write down its equation and its four axis intercepts.

Plan. Square the radius, then set one coordinate to zero at each axis.

x2+y2=62=36radius squaredy=0 ⟹ x=±6horizontal-axis interceptsx=0 ⟹ y=±6vertical-axis intercepts\begin{aligned} x^2+y^2&=6^2=36 &&\text{radius squared}\\ y=0&\ \Longrightarrow\ x=\pm6 &&\text{horizontal-axis intercepts}\\ x=0&\ \Longrightarrow\ y=\pm6 &&\text{vertical-axis intercepts} \end{aligned}

Why this step? On an axis, one coordinate is zero. The other must be one radius away from the origin in either direction.

Answer. x2+y2=36x^2+y^2=36, with intercepts (6,0),(−6,0),(0,6),(0,−6)(6,0),(-6,0),(0,6),(0,-6).

Check. Each intercept has coordinate squares summing to 3636 and distance 66 from the centre.

Change the direction of the question: the equation is given, and you must recover the radius before sketching.

Worked example 2: reading the radius

Question. Non-calculator. A circle has equation x2+y2=49x^2+y^2=49. Write down its centre and radius, and describe the points to mark for a sketch.

Plan. Compare with x2+y2=r2x^2+y^2=r^2, then take the positive square root.

r2=49compare equationsr=7positive square root(x,y)=(±7,0) or (0,±7)axis intercepts\begin{aligned} r^2&=49 &&\text{compare equations}\\ r&=7 &&\text{positive square root}\\ (x,y)&=(\pm7,0)\text{ or }(0,\pm7) &&\text{axis intercepts} \end{aligned}

Why this step? The 4949 is not the radius. It is the square of the distance to the centre.

Answer. Centre (0,0)(0,0), radius 77. Mark (7,0),(−7,0),(0,7),(0,−7)(7,0),(-7,0),(0,7),(0,-7), then draw a smooth circle on equal-scale axes.

Check. 72+02=497^2+0^2=49, while a supposed radius 4949 would give 49249^2, not 4949, on the right.

The equation contains both positive and negative coordinates. A circle is not just its first-quadrant arc.

Watch out

Wrong: a circle of radius 66 has equation x2+y2=6x^2+y^2=6. Right: x2+y2=36x^2+y^2=36. Also, x2+y2=0x^2+y^2=0 gives only the origin, not a circle with a positive radius.

Recover a radius from another equation.

Using coordinates on a circle

Substitution decides whether a point lies on a circle. A drawing can suggest an answer, but the equation settles it exactly.

For the circle x2+y2=25x^2+y^2=25, the point (3,4)(3,4) works because 32+42=253^2+4^2=25. Swapping the coordinates or changing either sign leaves the squares unchanged. In fact, exactly twelve points on this circle have integer coordinates: the four axis points and the eight sign-and-order variations of (3,4)(3,4). A continuous curve contains only a handful of these lattice points.

Worked example 3: checking a point

Question. Non-calculator. Does (−3,4)(-3,4) lie on the circle x2+y2=25x^2+y^2=25? Show working to support your answer.

Plan. Substitute both coordinates, including brackets around the negative one.

x2+y2=(−3)2+42substitute coordinates=9+16=25square then add\begin{aligned} x^2+y^2&=(-3)^2+4^2 &&\text{substitute coordinates}\\ &=9+16=25 &&\text{square then add} \end{aligned}

Why this step? The whole coordinate −3-3 is squared. Writing −32=−9-3^2=-9 would square only the 33 before applying the minus sign.

Answer. Yes. The coordinates satisfy x2+y2=25x^2+y^2=25, so the point lies on the circle.

Check. Reflecting (3,4)(3,4) in the yy-axis gives (−3,4)(-3,4) at the same distance from the origin.

Now change from checking two known coordinates to finding one missing coordinate. A vertical line can meet a circle twice, so keep both square roots unless a condition selects one.

Continue this lesson.

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