Construct both tangents from an external point

Try

A straight cable from an external anchor must just touch a circular pulley. Finding the contact point is a construction problem, not merely a length calculation. Let the circle have center O and positive radius r, and let the anchor P satisfy OP>r. A tangent contact T must make OT perpendicular to PT. That right-angle condition suggests an auxiliary circle.

As a separate worked example, take O=(0,0), r=4 and P=(10,0). The midpoint of OP is M=(5,0). Draw the circle centered at M with radius 5, so OP is its diameter. Its intersections with the radius-4 circle will be the two contacts. These values are separate from the explorer’s bounded presets.

Predict

Before constructing, predict the number of tangent lines when P is outside, on, or inside the circle. Decide why joining P to a contact at P itself cannot define a line. Predict what feature of the diameter circle guarantees the right angle at each external contact, and why centering the auxiliary circle at P would not achieve the same condition.

Experiment

Construct the contacts before drawing tangents

The given circle has center O=(0,0). Place P at the chosen distance and direction. Predict how many tangent lines pass through P, then advance the construction one step at a time. This is a worked construction explorer; your written practice must specify and justify the operations.

Circle radius23Distance OP0123456Direction of OPAlong positive xOblique direction (3/5,4/5)Along positive yAlong negative xReset tangent construction

1. Current: Given circle and point
2. Construct midpoint M of OP
3. Draw the circle with diameter OP
4. Locate the circle intersections
5. Draw and justify tangent lines

Previous construction stepNext construction stepStep 1 of 5

Solid circle: given. Dashed circle: auxiliary construction. Dotted segment: OP. Dashed radii and solid tangent lines appear in later steps. Lines continue beyond the fixed drawing window.

Construction coordinates (rounded where needed)

O	(0,0)
P	(5,0)

Compare an external, on-circle and interior point. Rotate OP to the oblique direction and explain why the argument still applies. Changing a given resets the construction; no animation, saved record or mastery credit is generated by these controls.

Choose the radius, point separation and direction, then predict the contact count before revealing the stages. Inspect the original data, midpoint, diameter circle, contacts and tangent lines in order. Record the right-angle condition and the incidence conditions separately. Displayed coordinates and dot products are approximate numerical checks; the theorem below proves the exact construction.

Test an oblique placement as well as an axial one. Change the separation through the boundary where it equals the radius. A coincident point pair is not a segment defining a tangent direction. Use the special perpendicular construction in that case rather than pretending that the two-contact procedure remains unchanged.

Understand

The executable external construction has four steps: construct the midpoint M of OP using its perpendicular bisector; draw the circle centered at M through O; mark its two intersections T₁,T₂ with the original circle; draw the full lines PT₁ and PT₂. The auxiliary circle also passes through P because MO=MP.

Every angle subtending diameter OP is right, so angle OTᵢP=90 degrees. Therefore PTᵢ is perpendicular to the corresponding radius and tangent. To prove completeness, start instead with any possible tangent contact T. Its radius is perpendicular to PT, so T satisfies the right-angle condition and lies on the circle with diameter OP by the converse diameter theorem. It also lies on the original circle. Hence every contact is among the intersections already constructed. For the converse diameter theorem, place O=(0,0),P=(d,0): the right-angle condition is x(d−x)−y²=0, equivalent to (x−d/2)²+y²=(d/2)². This independently proves the needed locus implication.

For the fixed worked example, subtract the circle equations x²+y²=16 and (x−5)²+y²=25. The second expands to x²+y²−10x=0, so x=8/5. Then y²=16−64/25=336/25, giving contacts (8/5,±4√21/5). Their tangent segment length is √(10²−4²)=2√21. The radius-contact dot product with P−T is x(10−x)−y²=10x−16=0, an independent arithmetic check.

More generally, with P=(d,0), externality d>r>0 gives x=r²/d and y²=r²(1−r²/d²)>0. Thus there are two distinct intersections. Rotating the configuration does not change this count. This coordinate calculation proves existence and supports the geometric completeness argument; it is not an instruction to measure and round a contact before drawing.

If d=r, the point is on the circle and there is one tangent, the line through P perpendicular to OP. The diameter circle meets the original only at P; the instruction “join P to P” fails, so construct the perpendicular explicitly. If d<r, every line through P has distance from O less than r and cuts the circle twice, giving no tangent. At d=0 the auxiliary diameter collapses as well. A radius-zero point locus is outside the positive-radius tangent theory.

Apply

Explain the difference between proving two candidate lines are tangent and proving there are no others. The first uses the diameter theorem; the second uses its converse and the entire circle-intersection set. In a construction report include the externality check, midpoint operation, both selected intersections and a reason for the count. A screen trace alone is not the general proof.

Continue

Next derive general circle equations and prove that every positive-radius circle is similar to every other. Both tasks require checking the reverse direction of a locus claim, just as completeness mattered here.