7.01f, 7.02b - Circles and perpendicular tangents

7.01f, 7.02b - Circles and perpendicular tangents

A circle equation describes every point the same distance from its centre. In 7.01f you will use an origin-centred circle, and in 7.02b you will find perpendicular lines and the tangent at a given point. These are Higher-tier coordinate-geometry skills.

The equation of a circle

Can you find the tangent to a circle at (3,4)(3,4) without drawing a careful tangent? The radius gives the direction you need, but first we must check that the point really lies on the circle.

The point (3,4)(3,4) is 55 units from the origin: a right triangle gives 32+42=523^2+4^2=5^2. Every point on the circle of radius 55 has that same distance. Negative coordinates do not change the squared distance.

[DIAGRAM: asset_name: A circle from Pythagoras; asset_slug: origin-circle; caption: Squared horizontal and vertical distances add to the squared radius.; description: An equal-scale coordinate diagram shows the circle x squared plus y squared equals twenty-five, the radius from the origin to three comma four, and a right triangle with horizontal length three and vertical length four. The four axis intercepts are marked.]
Squared horizontal and vertical distances add to the squared radius.

Origin-centred circle

x2+y2=r2,r>0x^2+y^2=r^2,\qquad r>0

The centre is (0,0)(0,0) and the radius is rr. This follows from Pythagoras. Not on the 2027 formula sheet: learn it.

To test a point, substitute both coordinates. Equality means on the circle. A smaller sum means inside and a larger sum means outside, because the sum is the square of the distance from the centre. The equation with right side zero gives only the origin; a negative right side gives no real points.

Worked example 1: identifying a circle

Question. State the centre, radius and axis intercepts of x2+y2=25x^2+y^2=25. Check whether (3,4)(3,4) lies on it.

Plan. Read the squared radius, then set one coordinate to zero for each pair of intercepts.

r=25=5positive radiusy=0  ⟹  x2=25  ⟹  x=±5horizontal interceptsx=0  ⟹  y2=25  ⟹  y=±5vertical intercepts32+42=9+16=25point test\begin{aligned} r&=\sqrt{25}=5 &&\text{positive radius}\\ y=0&\implies x^2=25\implies x=\pm5 &&\text{horizontal intercepts}\\ x=0&\implies y^2=25\implies y=\pm5 &&\text{vertical intercepts}\\ 3^2+4^2&=9+16=25 &&\text{point test} \end{aligned}

Answer. Centre (0,0)(0,0), radius 55, intercepts (5,0)(5,0), (−5,0)(-5,0), (0,5)(0,5), (0,−5)(0,-5). The point (3,4)(3,4) is on the circle.

Check. All four intercepts are distance 55 from the origin, matching the radius.

Change only the right-hand constant. The centre stays fixed, while the radius changes by its square root.

Worked example 2: a larger squared radius

Question. State the centre, radius and axis intercepts of x2+y2=100x^2+y^2=100. Decide whether the same point (3,4)(3,4) lies on it.

Plan. Repeat the radius and intercept calculations, then compare the point's squared distance.

r=100=10radius, not squared radiusy=0  ⟹  x=±10horizontal interceptsx=0  ⟹  y=±10vertical intercepts32+42=25<100distance is too small\begin{aligned} r&=\sqrt{100}=10 &&\text{radius, not squared radius}\\ y=0&\implies x=\pm10 &&\text{horizontal intercepts}\\ x=0&\implies y=\pm10 &&\text{vertical intercepts}\\ 3^2+4^2&=25<100 &&\text{distance is too small} \end{aligned}

Answer. Centre (0,0)(0,0), radius 1010, intercepts (±10,0)(\pm10,0) and (0,±10)(0,\pm10). The point (3,4)(3,4) is inside, not on, this circle.

Check. Its distance is still 55, which is smaller than the new radius 1010. You cannot find “the tangent at (3,4)(3,4)” to this circle because the point is not on it.

The equation is a distance statement, not a recipe that makes every supplied point a circle point.

Watch out

In x2+y2=49x^2+y^2=49, the radius is 77, not 4949. Take the positive square root of the constant.

Check both the radius and the proposed point.

Perpendicular gradients

Perpendicular lines meet at a right angle, and their directions are linked by a quarter turn.

A line of gradient 2/32/3 can move 33 units right and 22 up. Turn that direction through 90∘90^\circ: one perpendicular direction moves 22 right and 33 down, giving gradient −3/2-3/2. The rise and run exchange jobs, and one changes sign.

[DIAGRAM: asset_name: A quarter turn of direction; asset_slug: perpendicular-directions; caption: A right-three up-two direction turns into a right-two down-three direction.; description: Equal-scale axes show two perpendicular lines through the origin, with direction steps three comma two and two comma negative three. Dashed rise-and-run triangles label both gradients, and a right-angle marker lies between the directions.]
A right-three up-two direction turns into a right-two down-three direction.

Perpendicular finite gradients

m2=−1m1,m1m2=−1m_2=-\frac1{m_1},\qquad m_1m_2=-1

These forms apply when both gradients are finite and nonzero. A horizontal line has gradient zero and is perpendicular to a vertical line, whose gradient is undefined. Not on the 2027 formula sheet: learn these direction rules.

Once you know a finite gradient, substitute a point into y=mx+cy=mx+c to find cc. The perpendicular gradient alone does not choose one particular line.

Worked example 3: perpendicular to a rising line

Question. Find the line through (4,1)(4,1) perpendicular to y=32x+2y=\frac32x+2.

Plan. Take the negative reciprocal of the gradient, then use the point.

m=−23negative reciprocal of 3/21=−23(4)+csubstitute (4,1)c=113rearrange\begin{aligned} m&=-\tfrac23 &&\text{negative reciprocal of }3/2\\ 1&=-\tfrac23(4)+c &&\text{substitute }(4,1)\\ c&=\tfrac{11}{3} &&\text{rearrange} \end{aligned}

Answer. y=−23x+113y=-\frac23x+\frac{11}{3}.

Check. (3/2)(−2/3)=−1(3/2)(-2/3)=-1, and at x=4x=4 the new equation gives y=1y=1.

Change only the sign of the original gradient. Its perpendicular now rises.

Worked example 4: perpendicular to a falling line

Question. Find the line through (4,1)(4,1) perpendicular to y=−32x+2y=-\frac32x+2.

Plan. Use the new negative reciprocal and the same point.

m=23negative reciprocal of −3/21=23(4)+cpoint on the new linec=−53subtract 8/3\begin{aligned} m&=\tfrac23 &&\text{negative reciprocal of }-3/2\\ 1&=\tfrac23(4)+c &&\text{point on the new line}\\ c&=-\tfrac53 &&\text{subtract }8/3 \end{aligned}

Answer. y=23x−53y=\frac23x-\frac53.

Check. (−3/2)(2/3)=−1(-3/2)(2/3)=-1, and 8/3−5/3=18/3-5/3=1, so both conditions hold.

Horizontal and vertical lines need a direction argument rather than division by zero.

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