True or False? The angle in a semicircle theorem can be used on a diagram that shows a whole circle, not just one showing a semicircle.

True. All the theorem needs is a diameter with a triangle drawn on it, and a diameter can sit inside a whole circle just as easily. The rest of the circle plays no part.

True or False? The radius has to be drawn to the exact point where the tangent touches the circle for the right angle to appear.

True. A radius drawn to any other point on the circumference will meet the tangent at some other angle, or will not reach it at all.

How do the lengths of two tangents to a circle from one external point compare?

They are equal in length . The circle's symmetry guarantees this, however far the external point lies from the circle.

True or False? The line from the centre of a circle to an external point bisects the angle between the two tangents drawn from that point.

True. The whole figure is symmetrical about that line, so it splits the angle between the two tangents into two equal halves.

Two tangents from an external point, together with the radii to their points of contact, form two triangles. Why are these two triangles congruent?

They match by right angle, hypotenuse, side . Each has a right angle where its tangent meets its radius, the two radii are equal, and the line from the centre to the external point is a shared hypotenuse. The four sides together make a kite with its line of symmetry along that shared hypotenuse.

IGCSEMathsCambridge (CIE)International MathsCoreFlashcardsGeometryCircle Theorems

Circle Theorems (Cambridge (CIE) IGCSE International Maths: Core): Flashcards

Exam code: 0607

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FrontAngle in a Semicircle
A triangle is drawn in a circle so that one side is a diameter and the third vertex is on the circumference. What can you say about its angles?

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Cards in this collection (15)

A triangle is drawn in a circle so that one side is a diameter and the third vertex is on the circumference. What can you say about its angles?
The angle at the third vertex, opposite the diameter, is $90 °$ .
This result is called the angle in a semicircle, because only half of the circle is used.
True or False?
The angle in a semicircle is $90 °$ only when the point on the circumference is halfway along the arc.
False.
The angle is $90 °$ for every position of that point, so the vertex can sit anywhere on the arc between the two ends of the diameter.
Why does the angle in a semicircle theorem need a diameter rather than any chord?
Because a diameter passes through the centre, it splits the circle into two semicircles, with the vertex lying on one of them.
A chord that misses the centre cuts off a segment that is not a semicircle, and the angle there is not $90 °$ .
True or False?
The angle in a semicircle theorem can be used on a diagram that shows a whole circle, not just one showing a semicircle.
True.
All the theorem needs is a diameter with a triangle drawn on it, and a diameter can sit inside a whole circle just as easily.
The rest of the circle plays no part.
In a triangle whose side is a diameter, what do the two angles at the ends of the diameter add up to?
They add up to $90 °$ .
The right angle opposite the diameter uses up half of the triangle's $180 °$ , leaving the other two angles to share the rest.
What angle does a tangent make with the radius at its point of contact?
Exactly $90 °$ , so the tangent and that radius are perpendicular.
This holds at every point of the circle, since a tangent can be drawn at any point on the circumference.
True or False?
A tangent meets any chord of a circle at $90 °$ .
False.
The right angle needs a radius, which passes through the centre, so a chord that misses the centre does not meet the tangent at $90 °$ .
A circle has centre $O$ , and $P$ and $Q$ are two points on its circumference. Fill in the two gaps.
Since $O P$ and $O Q$ are both $______$ of the circle, triangle $O P Q$ is $______$ and its base angles are equal.
The completed sentence is:
Since $O P$ and $O Q$ are both radii of the circle, triangle $O P Q$ is isosceles and its base angles are equal.
Any two radii of the same circle are the same length, so this works for every such triangle.
True or False?
The radius has to be drawn to the exact point where the tangent touches the circle for the right angle to appear.
True.
A radius drawn to any other point on the circumference will meet the tangent at some other angle, or will not reach it at all.
A chord and a tangent meet at a point on a circle. How do you find the angle between that chord and the radius?
Subtract the angle between the chord and the tangent from $90 °$ .
The chord lies inside the right angle that the tangent and the radius already make, so it splits that $90 °$ into two parts.
How do the lengths of two tangents to a circle from one external point compare?
They are equal in length.
The circle's symmetry guarantees this, however far the external point lies from the circle.
True or False?
The line from the centre of a circle to an external point bisects the angle between the two tangents drawn from that point.
True.
The whole figure is symmetrical about that line, so it splits the angle between the two tangents into two equal halves.
Two tangents from an external point, together with the radii to their points of contact, form two triangles. Why are these two triangles congruent?
They match by right angle, hypotenuse, side.
Each has a right angle where its tangent meets its radius, the two radii are equal, and the line from the centre to the external point is a shared hypotenuse.
The four sides together make a kite with its line of symmetry along that shared hypotenuse.
A tangent from an external point $T$ touches a circle at $P$ , where the circle has centre $O$ and radius $r$ . Fill in the gap.
$P T = \sqrt{{(O T)}^{2} -______}$
The completed formula is:
$P T = \sqrt{{(O T)}^{2} - r^{2}}$
This is Pythagoras' theorem applied to triangle $O P T$ , in which $O T$ is the hypotenuse.
In the quadrilateral formed by two tangents and the two radii to their points of contact, how are the angle at the centre and the angle between the tangents related?
They add up to $180 °$ .
The four angles of the quadrilateral total $360 °$ , and the two angles at the points of contact are right angles, which uses up the other $180 °$ .

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