Circles: questions with answers

Each side is made of two tangent lengths, one from each of its endpoints. So $AB = 5 + 7$, $BC = 7 + 8$ and $CA = 8 + 5$. Adding these, every tangent length appears TWICE, once in each of the two sides through its vertex. So the perimeter is $2(5 + 7 + 8) = 40$ cm. Answering $20$ counts each tangent length only once, giving the sum of the three lengths instead of the perimeter. $60$ and $24$ use the wrong multiplier or leave out a vertex.

Apply Theorem 10.2 at each vertex and add the tangent lengths. The eight tangent lengths pair off so that $AB + CD = BC + DA$ -- the two sums that come out equal are over OPPOSITE sides. So $AB = BC + DA - CD = 9 + 10 - 4 = 15$ cm. Pairing adjacent sides instead gives $5$ from $AB + BC = CD + DA$, a relation the tangent lengths do not produce; $13$ and $19$ just add two of the given sides.

Equal tangent lengths imply that the sums of opposite sides of a quadrilateral circumscribing a circle are equal. Let the adjacent side lengths of the parallelogram be $a$ and $b$. Its opposite sides are equal, so $2a = 2b$, giving $a = b$. All four sides therefore equal $12$ cm, and the perimeter is $4 \times 12 = 48$ cm. Treating the shape as a rectangle with an unknown second side incorrectly suggests that the perimeter cannot be determined. $36$ counts only three sides, while $96$ counts every side twice.

Case 1 of Section 10.3: there is no tangent through a point lying inside a circle. Every line drawn through an interior point crosses the circle at two points, so it is a secant, not a tangent. Two is the count for an external point, and giving it here is answering the practised case; one is the count for a point on the circle. Where the point lies is the whole question.

Activity 3 settles the count by where the point lies. Through a point ON the circle there is exactly ONE tangent. Activity 1 rotated a wire about such a point and found that every position but one cuts the circle again. Two is the answer for a point OUTSIDE the circle, and giving it here is answering the practised case rather than this one. Zero belongs to a point inside, and 'infinitely many' confuses the tangents AT one point with the tangents at all the points of the circle.

The two parallel tangents touch at opposite ends of the diameter perpendicular to them. So the distance between them is the diameter itself, $10$ cm. Here $5$ is the radius, the distance from the centre to each tangent, not the answer. $20$ and $40$ stretch the diameter by a factor nothing in the question calls for.

The radius $OT$ is perpendicular to the tangent at the point of contact, so triangle $OTP$ is right-angled at $T$. Its other two angles add to $90^\circ$, so $\angle POT = 90^\circ - 50^\circ = 40^\circ$. The option $130^\circ$ wrongly adds the two angles to $180^\circ$. The option $90^\circ$ wrongly places the right angle at the centre instead of at the point of contact.

The radius $OB$ is perpendicular to the tangent at $B$ (Theorem 10.1), so triangle $OBA$ has a right angle at $B$ and its other two angles add to $90^\circ$: $\angle OAB = 90^\circ - 25^\circ = 65^\circ$. $155^\circ$ comes from adding the two angles to $180^\circ$ instead of $90^\circ$. $90^\circ$ comes from placing the right angle at $A$ instead of at $B$.

$OM$ and $ON$ are both radii of the same circle, so triangle $OMN$ is ISOSCELES and its two base angles $\angle OMN$ and $\angle ONM$ are EQUAL. The three angles add to $180^\circ$, so each base angle is $\frac{180^\circ - 80^\circ}{2} = 50^\circ$. Answering $100^\circ$ takes the whole leftover angle without splitting it between the two equal base angles.

Tangent segments from the same external point are equal, so $CA = CE$, $DB = DE$, and $PA = PB$. Since $E$ lies between $C$ and $D$, $CD = CE + ED = CA + DB$. Thus the perimeter is $(PC + CA) + (DB + DP) = PA + PB$. Therefore, the perimeter is $2 \times 16 = 32$ cm. Equal tangent lengths do not require a symmetric triangle.

$OP$ BISECTS the angle between the tangents (the Remark after Theorem 10.2), so each of the two congruent right triangles carries HALF of $60^\circ$, that is $30^\circ$, at $P$. In that triangle the radius $12$ is opposite the $30^\circ$ and the tangent is adjacent to it, so tangent $= 12 \cot 30^\circ = 12\sqrt{3}$ cm. Using the whole $60^\circ$ gives $\frac{12}{\sqrt{3}}$, which is shorter than the radius and is the value for a $60^\circ$ half-angle, not a $30^\circ$ one; $24$ is $OP$ and $12$ is the radius itself.

Theorem 10.1 says a radius meets a tangent at right angles at the point of contact. The points of contact here are $B$ and $C$, so the right angles of $OBAC$ are the ones at $B$ and $C$. The angle at $O$ and the angle at $A$ are not right angles in general. Naming all four angles wrongly treats the figure as a square.

The point of contact bisects the chord, so the right triangle contains HALF of it: $\frac{ 70 }{2} = 35$ cm. With the small radius $12$ cm as the other leg, the larger radius is the hypotenuse: $\sqrt{ 12^2 + 35^2 } = 37$ cm. Using the WHOLE chord as the leg gives $\sqrt{5044}$, which is the commonest slip here. $47$ adds the small radius to the half-chord as if they lay along one line, and $23$ subtracts them.

The perpendicular of $5$ cm from the centre bisects the chord. So half the chord is $\sqrt{ 13^2 - 5^2 } = 12$ cm, and the whole chord is $2 \times 12 = 24$ cm. Reporting $12$ stops at the leg the triangle actually contains, which is only half the chord. $26$ is the diameter, the longest chord of all, and $18$ adds the radius to the distance.

The chord touches the smaller circle, so the small radius drawn to the point of contact is perpendicular to it (Theorem 10.1). A perpendicular from the centre bisects a chord, so the point of contact is the midpoint of the chord (Example 1). Half the chord, the small radius $12$ and the large radius $37$ make a right triangle, so half the chord $= 35$ cm. The whole chord is $2 \times 35 = 70$ cm. Stopping at $35$ reports only half of it. $49$ adds the two radii, which lie along a diameter, not across a right angle. $25$ subtracts them instead of using Pythagoras.