C5.6, C5.7; E5.6, E5.7 - Right angles and tangents in circles

C5.6, C5.7; E5.6, E5.7 - Right angles and tangents in circles

Join the ends of a diameter to a third point on the circle. Why is the angle at that third point always a right angle, even when the point moves? C5.6 and E5.6 give two circle facts about right angles; C5.7 and E5.7 add the equal lengths of tangents from one external point.

Semicircle angles

The crucial information is that the chord passes through the centre. A chord that merely looks long does not give the same conclusion.

Angle in a semicircle

If ABAB is a diameter and CC is any other point on the circumference, then ∠ACB=90∘\angle ACB=90^\circ. The angle has its vertex on the circumference and its arms join the two ends of the diameter.

Here is why. Let OO be the centre and join OCOC. The radii OAOA, OBOB and OCOC are equal. In the two isosceles triangles, write ∠OAC=∠ACO=α\angle OAC=\angle ACO=\alpha and ∠OBC=∠BCO=β\angle OBC=\angle BCO=\beta. Since A,O,BA,O,B lie in a straight line, the full triangle ACBACB has angles α\alpha, β\beta and α+β\alpha+\beta. Its angle sum gives 2α+2β=180∘2\alpha+2\beta=180^\circ, so ∠ACB=α+β=90∘\angle ACB=\alpha+\beta=90^\circ.

[DIAGRAM: asset_name: The right angle in a semicircle; asset_slug: semicircle_right_angle; caption: Both triangles use the same diameter AB; moving the third vertex keeps the angle at C equal to 90 degrees.; description: Two exact circle panels have a horizontal diameter AB through centre O and different circumference points C above it. Both angles ACB are marked with right-angle squares. One panel includes OC and equal-radius ticks to support the isosceles-triangle explanation.]
Both triangles use the same diameter AB

Worked example 1: finding the right angle

Question. ABAB is a diameter of a circle and CC is another point on its circumference. Find ∠ACB\angle ACB, giving a reason.

Plan. Match the angle's arms to the two ends of the diameter.

  1. The angle's vertex CC is on the circle.
  2. Its arms CACA and CBCB meet the endpoints of diameter ABAB.
  3. Therefore ∠ACB=90∘\angle ACB=90^\circ, by the angle in a semicircle.

Answer. 90∘90^\circ, because the angle in a semicircle is a right angle.

Check. The right angle is at CC, opposite the diameter. It is not the straight angle AOBAOB at the centre.

The right angle now becomes one step in finding a different triangle angle.

Worked example 2: a second angle in the triangle

Question. PQPQ is a diameter and RR lies on the circumference. Given ∠RPQ=34∘\angle RPQ=34^\circ, calculate ∠PQR\angle PQR, giving reasons.

Plan. Establish the right angle at RR, then use the triangle angle sum.

∠PRQ=90∘angle in a semicircle∠PQR=180∘−90∘−34∘angles in a triangle=56∘\begin{aligned} \angle PRQ&=90^\circ &&\text{angle in a semicircle}\\ \angle PQR&=180^\circ-90^\circ-34^\circ &&\text{angles in a triangle}\\ &=56^\circ \end{aligned}

Answer. ∠PQR=56∘\angle PQR=56^\circ.

Check. The two acute angles are complementary: 34+56=9034+56=90. Together with the right angle they make 180∘180^\circ.

The point may be above or below the diameter; the condition is still the same. It must be on the circumference and different from the diameter's endpoints.

Watch out

“Angle in a semicircle” requires a diameter. A general chord does not guarantee 90∘90^\circ, and an angle with vertex at the centre is a different angle.

Identify the diameter before using the triangle sum.

Radius and tangent

A tangent touches a circle at one point. Draw the radius to that exact point of contact: it makes a right angle with the tangent on both sides.

Radius perpendicular to tangent

The tangent to a circle at TT is perpendicular to the radius OTOT, where OO is the centre. Thus the angle between OTOT and either direction along the tangent is 90∘90^\circ.

The radius points straight towards the tangent's contact point. Any other point on the tangent is outside the circle, so TT is the tangent's closest point to the centre. The shortest connection from a point to a line is perpendicular, explaining the right angle.

[DIAGRAM: asset_name: Radii and two tangents; asset_slug: radius_and_equal_tangents; caption: Each radius meets its tangent at 90 degrees, and the two tangent segments from P have equal lengths.; description: Exact circle with centre O, an external point P and contact points A and B. PA and PB touch the circle at A and B respectively, OA and OB have equal-radius ticks, and right-angle squares mark OAP and OBP. The central line OP is dashed to show the reflection symmetry exchanging the contact points.]
Each radius meets its tangent at 90 degrees, and the two tangent segments from P have equal lengths.

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