Question

Draw a circle of radius 3 cm. Take a point A on its extended diameter at a distance of 7 cm from its centre. Draw two tangents to the circle from A​

Answer 1

Step-by-step explanation:

Step 1: Draw a circle with centre O and radius 3 cm using a compass.

Step 2: Draw a secant passing through the centre. Mark points P and Q on opposite sides of the centre at a distance of 7 cm from O.

Step 3: Place the compass on P and draw two arcs on opposite sides of OP. Now place the compass on O and draw two arcs intersecting the arcs drawn from point P.

Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of OP. Mark the mid point of OP as M

1

Step 5: From M

1

draw a circle with radius =M

1

P=M

1

O

Step 6: Mark the intersection points of the circle drawn from M

1

with the circle drawn from O as A and B.

Step 7: Join P−A and P−B

Step 8: Repeat steps 3 to 7 for point Q and obtain tangents QC and QD

PA,PB,QC and QD are the required tangents.

Answer 2

Answer:

Steps of Construction:

(a) Bisect PO. Let M be the mid-point of PO.

(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points A and B.

(c) Join PA and PB. Then PA and PB are the required two tangents.

(d) Bisect QO. Let N be the mid-point of QO.(e) Taking N as centre and NO as radius, draw a circle. Let it intersects the given circle at the points C and D.

(f) Join QC and QD.

Then QC and QD are the required two tangents.

Justification:

Join OA and OB.

Then PAO is an angle in the semicircle and therefore ∠PAO = 90° .

PA ⊥ OA

Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.

Again join OC and OD.

Then ∠QCO is an angle in the semicircle and therefore ∠QCO = 90° .

Since OC is a radius of the given circle, QC has to be a tangent to the circle. Similarly, QD is also a tangent to the circle.

Step-by-step explanation:

Hope this helps you ✌️