WHEN CENTRE IS GIVEN:
When point of tangency is on the circle.
Given : A circle with centre O.
Required : To draw a tangent from point P on the circle.
Steps of construction:
1. Take a point O on the plane of the paper and draw a circle of given radius.
2. Take a point P on the circle.
3. Join OP.
4. Construct ∠OPX = 90°.
5. Produce XP to Y to get XPY as the required tangent.
When point of tangency is outside the circle
Given: A circle with centre O.
Required: To draw a tangent from an external point, i.e. P.
Steps of construction:
1. Join the centre O of the circle to the given external point, i.e. P.
2. Draw ⊥ bisector of OP, intersecting OP at O′.
3. Taking O′ as centre and OO′ = PO′ as radius, draw a circle to intersect the given circle at T and T′.
4. Join PT and PT′ to get the required tangents as PT and PT′.
WHEN CENTRE IS NOT GIVEN:
⇒ When point of tangency is on the circles.
Given: A circle and a point P on it.
Required: To draw a tangent at P without using centre of the circle.
Steps of construction:
1. Draw any chord PQ of the circle through P as in figure.
2. Take any point R on the major arc PQ and join PR and QR.
3. Construct ∠QPX equal to ∠PRQ.
Then PX is the required tangent at P to the circle.
When point of tangency is outside the circle.
Given: A circle and a point P outside it.
Required: To draw a tangent from point P without using the centre.
Steps of construction:
1. Let P be the external point from where the tangents are to be drawn to the given circle.
2. Through P draw a secant PAB to intersect the circle at A and B.
3. Produce AP to a point C such that AP = PC.
4. Draw a semi-circle with BC as diameter.
5. Draw PD ⊥ CB, intersecting the semi-circle at D.
6. With P as centre and PD as radius draw arcs to intersect the given circle at T and T′.
7. Joint PT and PT′. PT and PT′ are the required tangents.
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Given:
(i) A circle with centre O and radius 6 cm.
(ii) A point P at a distance of 10 cm from O.
Required: To draw two tangents to the circle from point P and measure the lengths of the tangents.
Steps of construction:
1. Bisect the line-segment OP. Let the point of bisection be M.
2. Taking M as centre and OM as radius, draw a circle. Let is intersect the circle at the points Q and R.
3. Join PQ and PR. These are the required tangents. On measuring, PQ = PR = 8 cm.
2. Draw tangents from an external point P to a circle of radius 4 cm without using the centre.
Solution:
Given: A circle of radius 4 cm.
Required: To draw two tangents from an external point P.
Steps of construction:
1. Draw a circle of radius 4 cm.
2. Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
3. Produce AP to C such that PA = CP.
4. Draw a semi-circle with CB as diameter.
5. Draw PD ⊥ CB, intersecting the semi-circle at D.
6. With P as centre and PD as radius draw arcs to intersect the given circle at T and T′.
7. Join PT and PT′. Then PT and PT′ are the required tangents.
3. Draw a line segment AB of length 6 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Sol.
Steps of construction:
1. Draw a line segment AB = 6 cm.
2. With centre A and radius = 3 cm, draw a circle.
3. With centre B and radius = 2 cm, draw another circle.
4. With M, the midpoint of AB, as centre and radius AM = MB, draw the circle intersecting the circle of radius 3 cm at R and S and intersecting the circle of radius 2 cm at P and Q.
5. Join AP, AQ, BR and BS.
Then AP and AQ are tangents to the circle with centre B and radius 2 cm from point A
BR and BS are tangents to the circle with centre A and radius 3 cm from point B.
Justification :
Join AR.
∠ARB = 90
o
[Angle in a semicircle]
i.e. BR is perpendicular to AR
Since AR is the radius of the circle with centre A and radius 3 cm, BR has to be a tangent to this circle.
Similarly BS is also a tangent this circle.
Join BP
∠BPA = 90
o
[Angle in a semicircle]
i.e. AP is perpendicular to BP.
Since BP is the radius of the circle with centre B and radius 2 cm, AP has to be tangent to this circle.
Similarly, AQ is also a tangent to this circle.
4. From a point P outside a circle of radius 2 cm, draw two tangents to the circle without using its centre.
Solution:
Given: A circle of radius 2 cm and a point P outside the circle.
Required: To draw two tangents from P to the circle without using its centre.
Steps of construction:
1. Draw a secant PAB to the circle.
2. Draw XY, the perpendicular bisector of PB. Let it intersect PB at M.
3. With M as centre and radius equal to MP or MB, draw a semicircle.
4. Draw a line perpendicular to PB through A. Let it intersect the semicircle at the point C.
5. With P is as centre and PC as radius, draw arcs to intersect the given circle at two points, say Q and R.
6. Join PQ and PR.
Then PQ and PR are the required tangents.
5. Draw a tangent from at point A on an arc.
Sol. Steps of Construction:
1. A is any point on arc XY.
2. Mark any two points B and C other than A on this arc.
3. Draw ⊥ bisectors of BC and AB which intersect each other at O.
4. Join OA. Draw ∠OAT = 90º.
5. PAT is the required tangent.
