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Grade 12 · Implicit circle tangents

Implicit circle tangents: A lower horizontal tangent

Implicit circle tangents: investigate a lower horizontal tangent with circle radius r = 1; final angle (degrees) = 270.

Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.

Watch the relationship

Paused
Implicit circle tangents: A lower horizontal tangent. x coordinate: 1. y coordinate: 0. Radius: 1. Slope −x/y: Vertical tangent; undefinedA radius is normal to its tangentxy0r = 1θ = 0°Equal axis scales · coordinates in the readouts
The circle is centered at zero and has positive radius. Equal drawing scales preserve perpendicularity. The tangent slope is reported as undefined when y is zero; the normal vector is the radius divided by r.

Starts paused. Play once, pause anywhere, or use Step to inspect the mathematics. Playback stops when this panel leaves the screen.

Make it your experiment

Change one value. Notice what follows.

The controls adjust the model. Numbers below describe the current frame. Decimals are rounded.

x coordinate
1
y coordinate
0
Radius
1
Slope −x/y
Vertical tangent; undefined

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From experiment to screen.

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Understand what you are seeing

The idea behind the motion.

An implicit equation describes both halves of a circle without choosing one as a function of x. Differentiating relates the tangent slope to the current coordinates. At y=0 the tangent is vertical, so the slope formula has no finite value even though the circle itself remains smooth. This investigation starts with Circle radius r = 1; Final angle (degrees) = 270. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.

A relationship to keep

x²+y²=r²; dy/dx=−x/y when y≠0

Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.

  1. STEP 1

    Set up the mathematical model

    Read the radius and verify the point satisfies x²+y²=r². The starting case is “A lower horizontal tangent.”

  2. STEP 2

    Follow the changing quantity

    Rotate the radius to the selected angle and compare it with the tangent, which is perpendicular to that radius.

  3. STEP 3

    Explain and test the result

    At a horizontal-axis crossing, explain why a vertical tangent is not a missing point of the circle. Distinguish an undefined dy/dx from a nonsmooth curve.

Your turn to explain

Make a prediction. Test your reasoning.

Keep Circle radius r = 1; Final angle (degrees) = 270. Pause the timeline at 60%. Given x coordinate = -0.951, calculate y coordinate, radius, slope −x/y. Show the substitution into the displayed formula.

Use the values specified in the question. Reset restores the initial values for this investigation.

Compare your explanation

The point is (1 cos(162°), 1 sin(162°))=(-0.951057,0.309017), with squared radius 1. Implicit differentiation gives dy/dx=−(-0.951057)/(0.309017)=3.077684. Results: y coordinate: 0.309; Radius: 1; Slope −x/y: 3.078. Decimal values are rounded; retain the original parameters when checking.

Work through a full lesson

Connect the animation to a worked example and practice questions.

Grade 12 · Calculus

Differentiate a relation without solving it into one global function

Use the chain rule when both coordinates vary along a curve.