Question

Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length f 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.

Answer 1

(I) Draw PQ, the perpendicular bisector of chord AC. PQ is the required locus, which is the diameter of the circle.
Reason: We know each point on the perpendicular bisector of AB is equidistant from A and B. Also the perpendicular bisector of a chord, passes through the centre of the circle and any chord passing through the centre of the circle is its diameter.

∴ PQ is the diameter of the circle.
(ii) Chords AB and AC intersects at M and N is a moving point such that LM = LN, where LM ⊥ AB and LN ⊥ AC
In right ΔALN and ΔALB
∠ANL = ∠ABL               ...(90° each)
AL = AL                           ...(Common)
NL = BL                          ...[Given]
∴ ΔALN = ΔALB             ...[R.H.S.]
Hence ∠MAL = ∠BAL     ...c.p.c.t.
Thus, L lies on the bisector of ∠BAC.
Hence proved.

Answer 2

Draw a circle of radius 4 cm whose centre is O. Take a point A on the circumference of this circle.
With A as centre and radius 6 cm draw an arc to cut the circumference at B. Join AB.

Then AB is the chord of the circle of length 6 cm.
With A as centre and radius 5 cm draw another arc to cut the circumference at C. Join AC then AC is the chord of the circle of length 5 cm.
With A as centre and a suitable radius, draw two arcs on opposite sides of AC.
With C as centre and the same radius, draw two arcs on opposite sides of AC to intersect the former arcs at P and Q.
Join PQ and produce to cut the circle at D and E.
Join DE. Then chord DE is the locus of points inside the circle that Ls equidistant from A and C.
As chord DE passes through (he centre O of the circle, it is a diameter. To prove the construction take any point S inside the circle on DE.