A Tangent Length and a Chord Distance: Right Answers and Wrong Reasons
Ms. Brennan is grading a circle unit quiz and finds that Priya has one fully wrong item and one item that is correct with faulty reasoning. On Item 3, circle O has radius 5 and point P lies outside the circle with OP = 13. A tangent from P touches the circle at T, and students must find PT. Priya writes $PT = \sqrt{13^2 + 5^2} = \sqrt{194} \approx 13.9$, explaining that "you add the squares to get the tangent." On Item 4, chord AB is 8 cm long in a circle of radius 5 cm, and Priya writes that the distance from the center to the chord is 3 cm because "5-4-3 is a right triangle." The answer 3 cm is correct, but her written explanation labels the radius as a leg and the distance as the hypotenuse. Ms. Brennan wants one round of feedback that fixes the triangle orientation in both items.
1. Give the correct value of PT and identify precisely where Priya's setup went wrong, using the tangent-radius relationship.
Read the worked answer
$PT = 12$, not $\sqrt{194}$.
A tangent line is perpendicular to the radius at the point of tangency, so $OT \perp PT$ and $\angle OTP = 90°$. Therefore triangle OTP is right with the right angle at T. The side opposite that right angle is OP, which must be the hypotenuse; OT is the radius, a leg.
$$PT^2 + 5^2 = 13^2 \implies PT^2 = 169 - 25 = 144 \implies PT = 12.$$
Priya added the two squares instead of subtracting, which is exactly what happens when a leg is mistaken for the hypotenuse. Her setup treats OP and OT as legs meeting at a right angle, but the perpendicular is at T, not at O. Naming the right angle's vertex before writing the Pythagorean equation would have prevented the error: the side opposite the right angle is always the hypotenuse.
2. Priya's answer of 3 cm for the chord item is right. Evaluate her reasoning and supply the correct explanation, including why the radius is not a leg here.
Read the worked answer
The number is correct, but the reasoning misidentifies the hypotenuse in the same way as Item 3 - so the feedback should credit the value and repair the justification.
Drop a perpendicular from O to chord AB and let M be the foot. A perpendicular from the center to a chord bisects it, so $AM = MB = 4$ cm. Triangle OMA is right at M, with OA the radius. The right angle is at M, so OA = 5 is the hypotenuse, AM = 4 is a leg, and OM is the other leg: $$OM = \sqrt{5^2 - 4^2} = \sqrt{9} = 3.$$
Priya's explanation calls the radius a leg and the distance the hypotenuse, which would force $d = \sqrt{5^2 + 4^2} = \sqrt{41}$, contradicting her own answer. The correct picture: the radius runs from the center to a point on the circle, so it is always the longest of the three segments and therefore the hypotenuse. The 3-4-5 triangle is real here, but the 5 is the radius.
3. Write a single feedback note that addresses both items together, and explain why pairing them is more useful than correcting each separately.
Read the worked answer
Pairing them targets one habit rather than two isolated answers.
Note: "Priya - in both problems you built a right triangle, which is exactly right. The slip is which side is the hypotenuse. A hypotenuse is opposite the right angle and is the longest side. In Item 3 the right angle is at T, so OP = 13 is the hypotenuse: $PT = \sqrt{13^2 - 5^2} = 12$. In Item 4 the right angle is where the perpendicular from O meets AB, so the radius 5 is the hypotenuse: $d = \sqrt{5^2 - 4^2} = 3$. Before applying the Pythagorean theorem, mark the right angle and circle the side across from it. Then you will always subtract."
Correcting separately would leave Priya thinking she has two unrelated mistakes. The paired note names a single transferable procedure - identify the right angle, then identify the hypotenuse - and it credits the correct 3 cm instead of burying it, which keeps the useful part of her work intact.