Today our topic of discussion is Constructions related to circles.
Constructions related to circles
Constructions related to circles
Construction 6. To determine the centre of a circle or an arc of a circle.
Given a circle as in figure (a) or an arc of a circle as in figure (b). It is required to determine the centre of the circle or the arc.
Drawing: In the given circle or the arc of the circle, three different points A, B and C are taken. Join A, B and B, C. The perpendicular bisectors EF and GH of the chords AB and BC are drawn respectively. Let the bisectors intersect at 0. Threfore, the O is the required centre of the circle or of the are of the circle.
(a): Circle
(b): Arc
Proof: By construction, the line segment EF is the perpendicular bisector of chord AB and GH is the perpendicular bisector of chord BC. But both EF and GH pass through the centre and their common point is O. Therefore, the point O is the centre of the circle or of the arc of the circle.
Tangents to a Circle
We have known that a tangent can not be drawn to a circle from a point internal to it. If the point is on the circle, a single tangent can be drawn at that point. The tangent is perpendicular to the radius drawn from the specified point.
Therefore, in order to construct a tangent to a circle at a point on it, it is required to construct the radius from the point and then construct a perpendicular to it. Again, if the point is located outside the circle, two tangents to the circle can be constructed.
Construction 7. To draw a tangent at any point of a circle. Let. A be a point of a circle whose centre is O. It is required to draw a tangent to the circle at the point A.
Drawing: O, A are joined. At the point A, a perpendicular AP is drawn to OA. Then AP is the required tangent.
Proof: The line segment OA is the radius passing through A and AP is perpendicular to it. Hence AP is the required tangent.
Note: At any point of a circle only one tangent can be drawn.
Construction 8.
To draw a tangent to a circle from a point outside.
Let P be a point outside of a circle whose centre is O. A tangent is to be drawn to the circle from the point P.
Drawing:
- Join P,O. The middle point. M of the line segment PO is determined.
- Now with M as centre and MO as radius, a circle is drawn. Let the new circle intersect the given circle at the points A and B.
- Join A, P and B, P. Then both AP, BP are the required tangents.
Proof: A, O and B, O are joined. PO is the diameter of the circle APB,
∴ ∠PAO 1 right angle = [the angle inscribed in the semi-circle is a right angle]
So the line segment OA is perpendicular to AP. Therefore, the line segment AP is a tangent at A to the circle with centre at O. Similarly, the line segment BP is also a tangent to the circle.
Nota Bene: Two and only two tangents can be drawn to a circle from an external point.
Construction 9.
To draw a circle circumscribing a given triangle.
Let ABC be a triangle. It is required to draw a circle circumscribing it. That is, a circle which passes through the three vertices A, B and C of the triangle is to be drawn.
Drawing:
- EM and FN, the perpendicular bisectors of AB and AC respectively are drawn. Let the line segments intersect each other at O.
- A, O are joined. With O as centre and radius equal to OA, a circle is drawn.
Then the circle will pass through the points A, B and C and this circle is the required circumcircle of ∠ABC.
Proof: B, and C, O are joined. The point O stands on EM, the perpendicular bisector of AB.
OA = OB
Similarly, OA = OC
∴ OA = OB = OC
Hence, the circle drawn with O as the centre and OA as the radius passes through the three points A, B and C. This circle is the required circumcircle of ∠ABC.
Work: In the above figure, the circumcircle of an acute angled-triangle is constructed. Construct the circumcircle of an obtuse and right-angled triangles.
Notice that for in obtuse-angled triangle, the circumcentre lies outside the triangle, in acute-angled triangle, the circumcentre lies within the triangle and in right-angled triangle, the circumcentre lies on the hypotenuse of the triangle.
Construction 10. To draw a circle inscribed in a triangle.
Let triangle ABC be a triangle. It is required to draw am inscribed circle. that is, to inscribe a circle in it or to draw a circle in it such that it touches each of the three sides BC, CA and AB of triangle ABC .
Drawing: BL and CM, the bisectors respectively of the angles ∠ABC and ∠ACB are drawn. Let the line segments intersect at O. OD is drawn perpendicular to BC from O and let it intersect BC B at D. With O as centre and OD as radius, a circle is drawn. Then, this circle is the required inscribed circle.
Proof: From O, OE and OF are drawn perpendiculars respectively to AC and AB. Let these two perpendiculars intersect the respective sides at E and F.
The point O lies on the bisector of angle ABC .
therefore OF = OD
Similarly, as O lies on bisector of ∠ACB, OE = OD
OD = OE = OF
Hence, the circle drawn with centre as O and OD as radius passes through D, E and F.
Again, OD, OE and OF respectively are perpendiculars to BC, AC and AB at their extremities.
Hence, the circle lying inside triangle ABC touches its sides at the points D, E and F respectively. Hence, the circle DEF is the required inscribed circle of ΔABC
Construction 11. To draw an ex-circle of a given triangle.
Let ABC be a triangle. It is required to draw its ex-circle. That is, to draw a circle which touches one side of triangle and the other two sides produced.
Drawing: Let AB and AC be produced to D and F respectively.
BM and CN, the bisectors of ∠ DBC and ∠FCB respectively are drawn. Let E be their point of intersection. From E_{i} perpendicular EH is drawn on BC and let EH intersects BC at H. With E as centre and radius equal to EH, a circle is drawn.
The circle HGL is the ex-circle of the ΔABC.
Proof: From E , perpendiculars EG and EL respectively are drawn to line segments BD and CF.Let the perpendicular intersect line segments BD and CF at the point G and L respectively.
Since E lies on the bisector of ∠DBC
∴ EH = EG
Similarly, the point E lies on the bisector of ∠FCB, so, EH = EL
∴ EH = EG = EL
Hence, the circle drawn with E as centre and radius equal to EL passes through H, G and L.
Again, the line segments BC, BD and CF respectively are perpendiculars at the extremities of EH, EG and EL. Hence, the circle touches the three line segments at the three points H, G and L.
Therefore, the circle HGL is the ex-circle of triangle ABC .
Remarks: Three ex-circles can be drawn with any triangle.
Work: Construct the two other ex-circles of a triangle.
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