Constructions (Circles) — Important Questions

Answer: (b) Perpendicular bisectors of the sides.

The circumcentre is equidistant from the vertices and is located at the intersection of the perpendicular bisectors of the sides.

Answer: (b) Internal bisectors of the angles.

The incentre is equidistant from the sides, so it lies where the internal angle bisectors meet.

Answer: (c) $2$.

Exactly two tangents can be drawn to a circle from an external point (from a point on the circle only one, and from an interior point none).

Answer: (c) $120^{\circ}$.

In quadrilateral $OAPB$, $\angle OAP=\angle OBP=90^{\circ}$, so $\angle AOB=360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}$.

Answer: (a) The construction draws the perpendicular to $OP$ at $P$ precisely because the tangent must be perpendicular to the radius at the point of contact; hence R correctly explains A.

Steps of construction:

1. Draw a circle with centre $O$ and radius $4\,\text{cm}$; mark any point $P$ on it.
2. Join $OP$ (the radius to the point of contact).
3. At $P$, construct a line perpendicular to $OP$ (using compasses to erect the perpendicular).

This perpendicular line is the required tangent, since a tangent is perpendicular to the radius at the point of contact.

In a right-angled triangle the hypotenuse subtends a right angle at the opposite vertex, so the hypotenuse is a diameter of the circumcircle (angle in a semicircle).

Hence the circumcentre is the midpoint of the hypotenuse $AC$, and the circumradius is
$R=\frac{AC}{2}=\frac{6}{2}=3\,\text{cm}.$

Steps of construction:

1. Draw $BC=6\,\text{cm}$. With centre $B$ and radius $5\,\text{cm}$, and centre $C$ and radius $7\,\text{cm}$, draw arcs meeting at $A$. Join $AB$ and $AC$ to complete $\triangle ABC$.
2. Construct the perpendicular bisector of $BC$ (equal arcs from $B$ and $C$, joined).
3. Construct the perpendicular bisector of $AC$ in the same way.
4. Let the two perpendicular bisectors meet at $O$; this is the circumcentre.
5. With centre $O$ and radius $OA$ (=,$OB=OC$) draw the circle; it passes through $A$, $B$, $C$ and is the required circumcircle.

Steps of construction:

1. Draw $QR=6\,\text{cm}$. At $Q$ construct $\angle Q=60^{\circ}$ and cut off $QP=5\,\text{cm}$; join $PR$ to complete $\triangle PQR$.
2. Construct the bisector of $\angle Q$.
3. Construct the bisector of $\angle R$. Let the two bisectors meet at $I$; this is the incentre.
4. From $I$, drop a perpendicular $IM$ to side $QR$.
5. With centre $I$ and radius $IM$, draw the circle; it touches all three sides and is the required incircle.

Steps of construction:

1. Draw a circle with centre $O$ and radius $3\,\text{cm}$. Mark $P$ with $OP=7\,\text{cm}$ and join $OP$.
2. Construct the perpendicular bisector of $OP$; let it meet $OP$ at its midpoint $M$.
3. With centre $M$ and radius $MO\ (=MP)$, draw a circle; it cuts the given circle at points $A$ and $B$.
4. Join $PA$ and $PB$. These are the required tangents, because $\angle OAP=\angle OBP=90^{\circ}$ (angles in the semicircle on diameter $OP$), so $PA\perp OA$ and $PB\perp OB$.

The length of each tangent $=\sqrt{OP^2-OA^2}=\sqrt{7^2-3^2}=\sqrt{40}=2\sqrt{10}\approx6.32\,\text{cm}.$

Steps of construction:

1. Draw $BC=7\,\text{cm}$. At $B$ construct $45^{\circ}$ and at $C$ construct $60^{\circ}$; the arms meet at $A$, completing $\triangle ABC$ (with $\angle A=75^{\circ}$).
2. Construct the perpendicular bisectors of $BC$ and $AB$; they meet at $O$, the circumcentre.
3. With centre $O$ and radius $OB$ draw the circumcircle through $A$, $B$, $C$.
4. Measure $OB$; on an accurate drawing the circumradius is about $4.3\,\text{cm}$ (using $R=\dfrac{a}{2\sin A}=\dfrac{7}{2\sin 75^{\circ}}\approx3.62\,\text{cm}$ as a check).

Reason for equidistance: Any point on the perpendicular bisector of a segment is equidistant from its two endpoints. $O$ lies on the perpendicular bisector of $BC$, so $OB=OC$; it also lies on the perpendicular bisector of $AB$, so $OA=OB$. Therefore $OA=OB=OC$, i.e. $O$ is equidistant from all three vertices, which is why a single circle passes through them.

Steps of construction:

1. Draw a circle with centre $O$ and radius $4\,\text{cm}$. Draw radii $OA$ and $OB$ with $\angle AOB=120^{\circ}$.
2. At $A$, construct the perpendicular to $OA$; at $B$, construct the perpendicular to $OB$. These are the tangents at $A$ and $B$.
3. Let the two tangents meet at $P$.

Finding $\angle APB$: In quadrilateral $OAPB$, $\angle OAP=\angle OBP=90^{\circ}$, so
$\angle APB=360^{\circ}-90^{\circ}-90^{\circ}-120^{\circ}=60^{\circ}.$

Finding $OP$: $OP$ bisects $\angle AOB$, so $\angle AOP=60^{\circ}$. In right triangle $OAP$ (right-angled at $A$),
$\cos 60^{\circ}=\frac{OA}{OP}\ \Rightarrow\ OP=\frac{OA}{\cos 60^{\circ}}=\frac{4}{\tfrac12}=8\,\text{cm}.$

Hence $\angle APB=60^{\circ}$ and $OP=8\,\text{cm}$.

(i) The incentre is located by drawing the internal bisectors of the angles of $\triangle ABC$ (any two suffice); their intersection is the centre. The radius is the perpendicular distance from that centre to a side.

(ii) Since $\angle B=90^{\circ}$, by Pythagoras
$AC=\sqrt{AB^2+BC^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\,\text{cm}.$

(iii) Area $=\tfrac12\times AB\times BC=\tfrac12\times6\times8=24\,\text{cm}^2$. Semi-perimeter $s=\dfrac{6+8+10}{2}=12\,\text{cm}$.
$r=\frac{\text{Area}}{s}=\frac{24}{12}=2\,\text{cm}.$

(iv) The designer must set the compasses to the inradius, $r=2\,\text{cm}$, with the point at the incentre.